(i) (9 cos α, 9 sin α)
Let P (h, k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 9 sin α
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus becomes x 2 + y 2 = 81
(ii) ( 9 cos α, 6 sin α)
Let P (h, k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 6 sin α
The locus of p(h, k) is obtained by replacing h by x and k by y
∴ The required locus is \(\frac{x^{2}}{81}+\frac{y^{2}}{36}\) = 1
(i) (9 cos α, 9 sin α)
Let P (h, k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 9 sin α
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus becomes x 2 + y 2 = 81
(ii) ( 9 cos α, 6 sin α)
Let P (h, k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 6 sin α
The locus of p(h, k) is obtained by replacing h by x and k by y
∴ The required locus is \(\frac{x^{2}}{81}+\frac{y^{2}}{36}\) = 1
(i) Two units from x-axis:
Let P (h, k) be any point on the required path. From the given data, we have k = 2
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is y = 2
(ii) Three units from y-axis:
Let (h, k) be any point on the required path. From the given data, we have h = 3.
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is x = 3
(i) Two units from x-axis:
Let P (h, k) be any point on the required path. From the given data, we have k = 2
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is y = 2
(ii) Three units from y-axis:
Let (h, k) be any point on the required path. From the given data, we have h = 3.
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is x = 3
The given moving points is (a cos 3 θ, a sin 3 θ)
The given moving points is (a cos 3 θ, a sin 3 θ)
Given P (-3, 1) lie on x 2 – 5x + ky = 0
⇒ (-3) 2 – 5(-3) + k(1) = 0
9 + 15 + k = 0 ⇒ k = -24
Q (2, b) lie on x 2 – 5x + ky = 0
(2) 2 – 5(2) + k(b) = 0 ⇒ 4 – 5(2) – 24b = 0
Given P (-3, 1) lie on x 2 – 5x + ky = 0
⇒ (-3) 2 – 5(-3) + k(1) = 0
9 + 15 + k = 0 ⇒ k = -24
Q (2, b) lie on x 2 – 5x + ky = 0
(2) 2 – 5(2) + k(b) = 0 ⇒ 4 – 5(2) – 24b = 0
Given A and B are the ends of the straight rod of length 8 unit on the x and y-axes. Let A be (a, 0) and B (0, b).
Let M (h, k) be the midpoint of AB (h,k) =
In the right-angled ∆ OAB
AB 2 = OA 2 + OB 2
8 2 = a 2 + b 2
64 = (2h) 2 + (2k)
64 = 4h 2 + 4k 2
h 2 + k 2 = \(\frac{64}{4}\) = 16
The locus of M (h, k) is obtained by replacing h by x and k by y
∴ The required locus is x 2 + y 2 = 16
Given A and B are the ends of the straight rod of length 8 unit on the x and y-axes. Let A be (a, 0) and B (0, b).
Let M (h, k) be the midpoint of AB (h,k) =
In the right-angled ∆ OAB
AB 2 = OA 2 + OB 2
8 2 = a 2 + b 2
64 = (2h) 2 + (2k)
64 = 4h 2 + 4k 2
h 2 + k 2 = \(\frac{64}{4}\) = 16
The locus of M (h, k) is obtained by replacing h by x and k by y
∴ The required locus is x 2 + y 2 = 16
Let P (h, k) be the moving point
Let the given point be A (3, 5) and B (1, -1)
We are given PA 2 + PB 2 = 20
⇒ (h – 3) 2 + (k – 5) 2 + (h – 1) 2 + (k + 1) 2 = 20
⇒ h 2 – 6h + 9 + k 2 – 10k + 25 + h 2 – 2h + 1 + k 2 + 2k + 1 = 20
(i.e.) 2h 2 + 2k 2 – 8h – 8k + 36 – 20 = 0
2h 2 + 2k 2 – 8h – 8k + 16 = 0
(÷ by 2 ) h 2 + k 2 – 4h – 4k + 8 = 0
So the locus of P is x 2 + y 2 – 4x – 4y + 8 = 0
Let P (h, k) be the moving point
Let the given point be A (3, 5) and B (1, -1)
We are given PA 2 + PB 2 = 20
⇒ (h – 3) 2 + (k – 5) 2 + (h – 1) 2 + (k + 1) 2 = 20
⇒ h 2 – 6h + 9 + k 2 – 10k + 25 + h 2 – 2h + 1 + k 2 + 2k + 1 = 20
(i.e.) 2h 2 + 2k 2 – 8h – 8k + 36 – 20 = 0
2h 2 + 2k 2 – 8h – 8k + 16 = 0
(÷ by 2 ) h 2 + k 2 – 4h – 4k + 8 = 0
So the locus of P is x 2 + y 2 – 4x – 4y + 8 = 0
Given A (1, – 6) and B (4, – 2).
Let P (h, k) be a point such that the line segment AB subtends a right angle at P.
∴ ∆ APB is a right-angled triangle.
AB 2 = AP 2 + BP 2 ………… (1)
AB 2 = (4 – 1) 2 + (- 2 + 6) 2
AB 2 = 3 2 + 4 2 = 9 + 16 = 25
AP 2 = (h – 1) 2 + (k + 6) 2
BP 2 = (h – 4) 2 + (k + 2) 2
(1) ⇒
25 = (h – 1) 2 + (k + 6) 2 + (h – 4) 2 + (k + 2) 2
25 = h 2 – 2h + 1 + k 2 + 12k + 36 + h 2 – 8h + 16 + k 2 + 4k + 4
25 = 2h 2 + 2k 2 – 10h + 16k + 57
2h 2 + 2k 2 – 10h + 16k + 57 – 25 = 0
2h 2 + 2k 2 – 10h + 16k + 32 = 0
h 2 + k 2 – 5h + 8k + 16 =0
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is x 2 + y 2 – 5x + 8y + 16 = 0
Given A (1, – 6) and B (4, – 2).
Let P (h, k) be a point such that the line segment AB subtends a right angle at P.
∴ ∆ APB is a right-angled triangle.
AB 2 = AP 2 + BP 2 ………… (1)
AB 2 = (4 – 1) 2 + (- 2 + 6) 2
AB 2 = 3 2 + 4 2 = 9 + 16 = 25
AP 2 = (h – 1) 2 + (k + 6) 2
BP 2 = (h – 4) 2 + (k + 2) 2
(1) ⇒
25 = (h – 1) 2 + (k + 6) 2 + (h – 4) 2 + (k + 2) 2
25 = h 2 – 2h + 1 + k 2 + 12k + 36 + h 2 – 8h + 16 + k 2 + 4k + 4
25 = 2h 2 + 2k 2 – 10h + 16k + 57
2h 2 + 2k 2 – 10h + 16k + 57 – 25 = 0
2h 2 + 2k 2 – 10h + 16k + 32 = 0
h 2 + k 2 – 5h + 8k + 16 =0
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is x 2 + y 2 – 5x + 8y + 16 = 0
Let the variable point R be (x, y). Let M (h, k) be the midpoint of R.
But R(x, y) is a point on y 2 = 4x
∴ (2k) 2 = 4(2h)
4k 2 = 8h
k 2 = 2h
The locus of M (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is y 2 = 2x
Let the variable point R be (x, y). Let M (h, k) be the midpoint of R.
But R(x, y) is a point on y 2 = 4x
∴ (2k) 2 = 4(2h)
4k 2 = 8h
k 2 = 2h
The locus of M (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is y 2 = 2x
Let the moving point P be (h, k)
By the given data we have
The locus of P ( h, k ) is obtained by replacing h by x and k by y
∴ The required locus is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
b 2 x 2 – a 2 y 2 = a 2 b 2
Let the moving point P be (h, k)
By the given data we have
The locus of P ( h, k ) is obtained by replacing h by x and k by y
∴ The required locus is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
b 2 x 2 – a 2 y 2 = a 2 b 2
Given P is (2, -7) and let Q be (x, y)
Given that Q is a point on 2x 2 + 9y 2 = 18
Let M (h, k) be the midpoint of PQ
2h = 2 + x, 2k = – 7 + y
x = 2h – 2, y = 2k + 7
But Q(x, y) is a point on 2x 2 + 9y 2 = 18
∴ 2 (2h – 2) 2 + 9 (2k + 7) 2 = 18
2 [4h 2 – 8h + 4] + 9 [4k 2 + 28k + 49] = 18
8h 2 – 16h + 8 + 36k 2 + 252k + 441 = 18
8h 2 + 36k 2 – 16h + 252k + 449 = 18
8h 2 + 36k 2 – 16h + 252k +431 =0
The locus of M ( h, k ) is obtained by replacing h by x and k by y.
∴ The required locus is
8x 2 + 36y 2 – 16x + 252y + 431 = 0
Given P is (2, -7) and let Q be (x, y)
Given that Q is a point on 2x 2 + 9y 2 = 18
Let M (h, k) be the midpoint of PQ
2h = 2 + x, 2k = – 7 + y
x = 2h – 2, y = 2k + 7
But Q(x, y) is a point on 2x 2 + 9y 2 = 18
∴ 2 (2h – 2) 2 + 9 (2k + 7) 2 = 18
2 [4h 2 – 8h + 4] + 9 [4k 2 + 28k + 49] = 18
8h 2 – 16h + 8 + 36k 2 + 252k + 441 = 18
8h 2 + 36k 2 – 16h + 252k + 449 = 18
8h 2 + 36k 2 – 16h + 252k +431 =0
The locus of M ( h, k ) is obtained by replacing h by x and k by y.
∴ The required locus is
8x 2 + 36y 2 – 16x + 252y + 431 = 0
Given R is any point on the x-axis and Q is any point on the y-axis.
Let R be (x, 0) and Q be (0, y)
Let P (h, k ) be the variable point on RQ such that RP = b and PQ = a
The point P ( h, k ) divides the line joining the points R(x, 0) and Q (0, y) in the ratio b: a
Substituting in equation (1), we have
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}\) = 1
Given R is any point on the x-axis and Q is any point on the y-axis.
Let R be (x, 0) and Q be (0, y)
Let P (h, k ) be the variable point on RQ such that RP = b and PQ = a
The point P ( h, k ) divides the line joining the points R(x, 0) and Q (0, y) in the ratio b: a
Substituting in equation (1), we have
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}\) = 1
Given P (6, 2), Q (-2, 1), R (a, b) are the vertices of ∆ PQR where R (a, b) lies on y = x 2 – 3x + 4
∴ b = a 2 – 3a + 4 (1)
Let the centroid of ∆ PQR be G (h, k)
Substituting in equation (1) we have
(1) ⇒ 3k – 3 = (3h – 4) 2 – 3(3h – 4) + 4
3k – 3 = 9h 2 – 24h + 16 – 9h + 12 + 4
9h 2 – 33h + 32 – 3k + 3 = 0
9h 2 – 33h – 3k + 35 = 0
The locus of G (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is 9x 2 – 33x – 3y + 35 = 0
Given P (6, 2), Q (-2, 1), R (a, b) are the vertices of ∆ PQR where R (a, b) lies on y = x 2 – 3x + 4
∴ b = a 2 – 3a + 4 (1)
Let the centroid of ∆ PQR be G (h, k)
Substituting in equation (1) we have
(1) ⇒ 3k – 3 = (3h – 4) 2 – 3(3h – 4) + 4
3k – 3 = 9h 2 – 24h + 16 – 9h + 12 + 4
9h 2 – 33h + 32 – 3k + 3 = 0
9h 2 – 33h – 3k + 35 = 0
The locus of G (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is 9x 2 – 33x – 3y + 35 = 0
Let Q be (a, b) lying on the locus
x 2 + y 2 + 4x – 3y + 7 = 0
∴ a 2 + b 2 + 4a – 3b + 7 = 0
Let the movable point P be (h, k)
Given P divides OQ externally in the ratio 3: 4
Substituting in equation (1) we have
h 2 + k 2 – 12h + 9k + 63 = 0
The locus of P(h, k) is obtained by replacing h by x and k by y.
∴ The required locus is
x 2 + y 2 – 12x + 9y + 63 = 0
Let Q be (a, b) lying on the locus
x 2 + y 2 + 4x – 3y + 7 = 0
∴ a 2 + b 2 + 4a – 3b + 7 = 0
Let the movable point P be (h, k)
Given P divides OQ externally in the ratio 3: 4
Substituting in equation (1) we have
h 2 + k 2 – 12h + 9k + 63 = 0
The locus of P(h, k) is obtained by replacing h by x and k by y.
∴ The required locus is
x 2 + y 2 – 12x + 9y + 63 = 0
Given that the required pointis3unitsfrom x-axis and 5 units from the point P (5, 1). Let Q (h, 3) and K (h,- 3) be the required points.
∴ PQ = 5
\(\sqrt{(5-\mathrm{h})^{2}+(1-3)^{2}}\) = 5
(5 – h) 2 + (- 2) 2 = 25
25 – 10h + h 2 + 4 = 25
h 2 – 10h + 29 – 25 = 0
h 2 – 10h + 4 = 0
PR = 5
(5 – k) 2 + 4 2 = 25
25 – 10k + k 2 + 16 = 25
k 2 – 10k + 16 = 0
∴ R (8, – 3), (2, – 3)
∴ Required points are
(5 + √21, 3), (5 – √21, 3), (8, – 3), (2, – 3)
Given that the required pointis3unitsfrom x-axis and 5 units from the point P (5, 1). Let Q (h, 3) and K (h,- 3) be the required points.
∴ PQ = 5
\(\sqrt{(5-\mathrm{h})^{2}+(1-3)^{2}}\) = 5
(5 – h) 2 + (- 2) 2 = 25
25 – 10h + h 2 + 4 = 25
h 2 – 10h + 29 – 25 = 0
h 2 – 10h + 4 = 0
PR = 5
(5 – k) 2 + 4 2 = 25
25 – 10k + k 2 + 16 = 25
k 2 – 10k + 16 = 0
∴ R (8, – 3), (2, – 3)
∴ Required points are
(5 + √21, 3), (5 – √21, 3), (8, – 3), (2, – 3)
Let A be (4, 0) and B be (-4, 0). Let the moving point be p(h, k)
Given PA + PB = 10
16h 2 + 200h + 625 = 25[h 2 + 8h + 16 + k 2 ]
16h 2 + 200h + 625 = 25h 2 + 200h + 400 + 25k 2
9h 2 + 26k 2 = 225
Let A be (4, 0) and B be (-4, 0). Let the moving point be p(h, k)
Given PA + PB = 10
16h 2 + 200h + 625 = 25[h 2 + 8h + 16 + k 2 ]
16h 2 + 200h + 625 = 25h 2 + 200h + 400 + 25k 2
9h 2 + 26k 2 = 225
(i) with y – intercept – 4
The equation the line with slope m and having y – intercept b is y = mx + b ——- (1)
Given b = – 4
∴ (1) ⇒ y = m x – 4
Given this line passes through the point (1, 1)
∴ 1 = m 1 – 4 ⇒ m = 5
∴ The required equation is y = 5x – 4
(ii) Slope m = 3, passing through (x 1, y 1 ) = (1, 1)
Equation of the line is y – y 1 = m(x – x 1 )
(i.e) y- 1 = 3(x – 1) ⇒ y – 1 = 3x – 3 ⇒ 3x – y = 2
(iii) (-2, 3)
The equation of line joining the two points (x 1, y 1 ) and (x 2, y 2 ) is
Given (x 1, y 1 ) = (1, 1), (x 2, y 2 ) = (- 2, 3)
∴ The equation of the required line is
– 3 (y – 1) = 2 (x – 1)
– 3y + 3 = 2x – 2
2x + 3y – 2 – 3 = 0
2x + 3y – 5 = 0
(iv) The perpendicular from the origin makes an angle 60° with x – axis
The equation of the line in the normal form is x cos α + y sin α = p ——- (1)
Given α = 60°
∴ cos 60° = \(\frac{1}{2}\), sin 60° = \(\frac{\sqrt{3}}{2}\)
(1) ⇒ x. \(\frac{1}{2}\) + y. \(\frac{\sqrt{3}}{2}\) = p
x + √3 = 2p —— (2)
This line passes through the point (1,1)
∴ 1 + √3 = 2p
Substituting for p in equation (2)
∴ The required equation is x + √3y = 1 + √3
(i) with y – intercept – 4
The equation the line with slope m and having y – intercept b is y = mx + b ——- (1)
Given b = – 4
∴ (1) ⇒ y = m x – 4
Given this line passes through the point (1, 1)
∴ 1 = m 1 – 4 ⇒ m = 5
∴ The required equation is y = 5x – 4
(ii) Slope m = 3, passing through (x 1, y 1 ) = (1, 1)
Equation of the line is y – y 1 = m(x – x 1 )
(i.e) y- 1 = 3(x – 1) ⇒ y – 1 = 3x – 3 ⇒ 3x – y = 2
(iii) (-2, 3)
The equation of line joining the two points (x 1, y 1 ) and (x 2, y 2 ) is
Given (x 1, y 1 ) = (1, 1), (x 2, y 2 ) = (- 2, 3)
∴ The equation of the required line is
– 3 (y – 1) = 2 (x – 1)
– 3y + 3 = 2x – 2
2x + 3y – 2 – 3 = 0
2x + 3y – 5 = 0
(iv) The perpendicular from the origin makes an angle 60° with x – axis
The equation of the line in the normal form is x cos α + y sin α = p ——- (1)
Given α = 60°
∴ cos 60° = \(\frac{1}{2}\), sin 60° = \(\frac{\sqrt{3}}{2}\)
(1) ⇒ x. \(\frac{1}{2}\) + y. \(\frac{\sqrt{3}}{2}\) = p
x + √3 = 2p —— (2)
This line passes through the point (1,1)
∴ 1 + √3 = 2p
Substituting for p in equation (2)
∴ The required equation is x + √3y = 1 + √3
Let AB be the line segment intercepted between the axes. Let A be (a, 0) and B be (0, b)
Given P (r, c) is the mid point of AB
The equation of the line AB having x-intercept a and y-intercept b is \(\frac{x}{a}+\frac{y}{b}\) = 1
Substituting for a, b in the above equation, we have
Let AB be the line segment intercepted between the axes. Let A be (a, 0) and B be (0, b)
Given P (r, c) is the mid point of AB
The equation of the line AB having x-intercept a and y-intercept b is \(\frac{x}{a}+\frac{y}{b}\) = 1
Substituting for a, b in the above equation, we have
Let the line divide the coordinate axis in the ratio 3: 10.
∴ x-intercept = 3k
y-intercept = 10k
∴ The equation of the straight line is \(\frac{x}{3 k}+\frac{y}{10 k}\) = 1
This line passes through the point P (1, 5).
∴ The required equation is
Let the line divide the coordinate axis in the ratio 3: 10.
∴ x-intercept = 3k
y-intercept = 10k
∴ The equation of the straight line is \(\frac{x}{3 k}+\frac{y}{10 k}\) = 1
This line passes through the point P (1, 5).
∴ The required equation is
The equation of the line with x – intercept a and y – intercept b is \(\frac{x}{a}+\frac{y}{b}\) = 1 ——- (1)
The length of the perpendicular from the origin (0, 0) to the line (1) is
The equation of the line with x – intercept a and y – intercept b is \(\frac{x}{a}+\frac{y}{b}\) = 1 ——- (1)
The length of the perpendicular from the origin (0, 0) to the line (1) is
(i) Choose Celsius degree along the x-axis and Fahrenheit degree along the y-axis.
Given a Freezing point in Celsius = 0°C
The freezing point in Fahrenheit degree = 32° F
∴ The Freezing point is (0°, 32°)
Also given Boiling point in Celsius = 100°C
The boiling point in Fahrenheit = 212° F
∴ The Boding point is (100°, 212°)
Let C denote the Celsius degree and F denote the Fahrenheit degree.
The equation of the path connecting the freezing point (0°, 32°) and the boiling point (100°, 212° ) is
which is the required relation connecting C and F.
(ii) To find the value of C for 98.6° F
(1) For 98.6° F To find C.
(iii) To find the value of F for 38° C,
For 38° C To find F
(i) Choose Celsius degree along the x-axis and Fahrenheit degree along the y-axis.
Given a Freezing point in Celsius = 0°C
The freezing point in Fahrenheit degree = 32° F
∴ The Freezing point is (0°, 32°)
Also given Boiling point in Celsius = 100°C
The boiling point in Fahrenheit = 212° F
∴ The Boding point is (100°, 212°)
Let C denote the Celsius degree and F denote the Fahrenheit degree.
The equation of the path connecting the freezing point (0°, 32°) and the boiling point (100°, 212° ) is
which is the required relation connecting C and F.
(ii) To find the value of C for 98.6° F
(1) For 98.6° F To find C.
(iii) To find the value of F for 38° C,
For 38° C To find F
Let us take the time T along the x-axis and the Distance D along the y-axis.
Given when time T = 15 s, the distance D = 1400 m
The corresponding point is (15, 1400)
Also when time T = 18 s, the distance D = 800 m.
The corresponding point is (18, 800)
(i) The distance between the place and the target:
∴ The relation connecting T and D is the equation of the straight line joining the points (15, 1400) and (18, 800)
To find the distance between the target and the place, Put T = 0 in equation (1)
4400 – D = 0 ⇒ D = 4400 m.
Required distance = 4400 m.
(ii) The distance covered by it in 15 seconds:
Put T = 15 in the above equation
15 = \(\frac{1400-\mathrm{D}}{200}\) + 15
∴ \(\frac{1400-\mathrm{D}}{200}\) = 0 ⇒ D = 1400 m.
(iii) Time taken to hit the Target:
When the target is reached D = 0
∴ (1) ⇒ T = \(\frac{1400-0}{200}\) + 15
T = \(\frac{1400}{200}\) + 15
T = 7 + 15 = 22 seconds
∴ The time taken to hit the target is 22 seconds
Let us take the time T along the x-axis and the Distance D along the y-axis.
Given when time T = 15 s, the distance D = 1400 m
The corresponding point is (15, 1400)
Also when time T = 18 s, the distance D = 800 m.
The corresponding point is (18, 800)
(i) The distance between the place and the target:
∴ The relation connecting T and D is the equation of the straight line joining the points (15, 1400) and (18, 800)
To find the distance between the target and the place, Put T = 0 in equation (1)
4400 – D = 0 ⇒ D = 4400 m.
Required distance = 4400 m.
(ii) The distance covered by it in 15 seconds:
Put T = 15 in the above equation
15 = \(\frac{1400-\mathrm{D}}{200}\) + 15
∴ \(\frac{1400-\mathrm{D}}{200}\) = 0 ⇒ D = 1400 m.
(iii) Time taken to hit the Target:
When the target is reached D = 0
∴ (1) ⇒ T = \(\frac{1400-0}{200}\) + 15
T = \(\frac{1400}{200}\) + 15
T = 7 + 15 = 22 seconds
∴ The time taken to hit the target is 22 seconds
Let us choose the year along the x-axis and the population of the city along the y-axis.
Given In the year 2005 population is 1,35,000
The corresponding point is (2005, 1,35,000)
In the year 2010, population is 1,45,000
The corresponding point is (2010, 1,45,000)
Let Y denote the year and P denote the population.
The relation connecting Y and P is the equation of the straight line joining the points (2005, 1,35,000) and (2010, 1,45,000)
When y = 2015
P = 1,35,000 + 10 × 2,000
P = 1,35,000 + 20,000 = 1,55,000
∴ The population in the year 2015 is 1,55,000
Let us choose the year along the x-axis and the population of the city along the y-axis.
Given In the year 2005 population is 1,35,000
The corresponding point is (2005, 1,35,000)
In the year 2010, population is 1,45,000
The corresponding point is (2010, 1,45,000)
Let Y denote the year and P denote the population.
The relation connecting Y and P is the equation of the straight line joining the points (2005, 1,35,000) and (2010, 1,45,000)
When y = 2015
P = 1,35,000 + 10 × 2,000
P = 1,35,000 + 20,000 = 1,55,000
∴ The population in the year 2015 is 1,55,000
Given length of the perpendicular p = 12
Angle made by the perpendicular α = 30°
The equation the straight line in the normal form is
x cos α + y sin α = p
∴ The required equation of the straight line is
x cos 30° + y sin 30° = 12
x\(\frac{\sqrt{3}}{2}\) + y × \(\frac{1}{2}\) = 12
√3x + y = 24
Given length of the perpendicular p = 12
Angle made by the perpendicular α = 30°
The equation the straight line in the normal form is
x cos α + y sin α = p
∴ The required equation of the straight line is
x cos 30° + y sin 30° = 12
x\(\frac{\sqrt{3}}{2}\) + y × \(\frac{1}{2}\) = 12
√3x + y = 24
Let a and b be the x and y-intercepts of the line.
Given a + b = 1
b = 1 – a —— (1)
The equation of the straight line is
The line passes through the point (8, 3)
8(1 – a) + 3a = a(1 – a)
8 – 8a + 3a = a – a 2
a 2 – 5a + 8 – a = 0
a 2 – 6a + 8 = 0
a 2 – 4a – 2a + 8 = 0
a(a – 4) – 2(a – 4) = 0
(a- 4) (a – 2) = 0
a = 4 or a = 2
When a = 2, b = 1 – 2 = – 1
When a = 4, b = 1 – 4 = – 3
∴ The equation of the straight lines are
x – 2y = 2 and 3x – 4y = 12
Let a and b be the x and y-intercepts of the line.
Given a + b = 1
b = 1 – a —— (1)
The equation of the straight line is
The line passes through the point (8, 3)
8(1 – a) + 3a = a(1 – a)
8 – 8a + 3a = a – a 2
a 2 – 5a + 8 – a = 0
a 2 – 6a + 8 = 0
a 2 – 4a – 2a + 8 = 0
a(a – 4) – 2(a – 4) = 0
(a- 4) (a – 2) = 0
a = 4 or a = 2
When a = 2, b = 1 – 2 = – 1
When a = 4, b = 1 – 4 = – 3
∴ The equation of the straight lines are
x – 2y = 2 and 3x – 4y = 12
Let the given points be A (1, 3), B (2, 1) and \(\left(\frac{1}{2}, 4\right)\)
(i) Slope Method:
A (1, 3 ), B (2, 1 ), C\(\left(\frac{1}{2}, 4\right)\)
From equations (1) and (2)
Slope of AB = Slope of BC
∴ The given points A, B, C are collinear.
(ii) Using a straight line
A(1, 3), B(2, 1), C\(\left(\frac{1}{2}, 4\right)\)
The equation of the straight line joining the points
A( 1, 3), B(2, 1) is
-2(x- 1) = y – 3
– 2x + 2 = y – 3
2x + y – 2 – 3 = 0
2x+ y – 5 = 0 ——- (2)
Substituting the third point C \(\left(\frac{1}{2}, 4\right)\) in equation (2)
we have 2\(\left(\frac{1}{2}\right)\) + 4 – 5 = 0
1 + 4 – 5 = 0
0 = 0
∴ The third point C\(\left(\frac{1}{2}, 4\right)\) lies on the straight line AB.
Hence the points A, B, C are collinear.
(iii) Distance method:
A(1, 3), B(2, 1), C\(\left(\frac{1}{2}, 4\right)\)
Thus BA + AC = BC
∴ The points A, B, C are collinear.
Let the given points be A (1, 3), B (2, 1) and \(\left(\frac{1}{2}, 4\right)\)
(i) Slope Method:
A (1, 3 ), B (2, 1 ), C\(\left(\frac{1}{2}, 4\right)\)
From equations (1) and (2)
Slope of AB = Slope of BC
∴ The given points A, B, C are collinear.
(ii) Using a straight line
A(1, 3), B(2, 1), C\(\left(\frac{1}{2}, 4\right)\)
The equation of the straight line joining the points
A( 1, 3), B(2, 1) is
-2(x- 1) = y – 3
– 2x + 2 = y – 3
2x + y – 2 – 3 = 0
2x+ y – 5 = 0 ——- (2)
Substituting the third point C \(\left(\frac{1}{2}, 4\right)\) in equation (2)
we have 2\(\left(\frac{1}{2}\right)\) + 4 – 5 = 0
1 + 4 – 5 = 0
0 = 0
∴ The third point C\(\left(\frac{1}{2}, 4\right)\) lies on the straight line AB.
Hence the points A, B, C are collinear.
(iii) Distance method:
A(1, 3), B(2, 1), C\(\left(\frac{1}{2}, 4\right)\)
Thus BA + AC = BC
∴ The points A, B, C are collinear.
Slope of the line m = tan θ = \(\frac{5}{12}\)
sin θ = \(\frac{5}{13}\), cos θ = \(\frac{12}{13}\)
The parametric equation of the line passing through the point (1, 2) making angle θ with x – axis is
Any point on this line is
(1 + r cos θ, 2 + r sin θ) ……… (1)
where r is the distance of any point from A (1, 2) on the line.
To find the point which is 13 units away from A (1, 2) on the line.
Substitute r = ± 13, cos θ = \(\frac{12}{13}\), sin θ = \(\frac{5}{13}\) in equation (1)
Required point = \(\left(1 \pm 13\left(\frac{12}{13}\right), 2 \pm 13\left(\frac{5}{13}\right)\right)\)
= (1 ± 12, 2 ± 5)
= (1 + 12, 2 + 5)
= (1 + 12, 2 + 5) or (1 – 12, 2 – 5)
= (13, 7) or (- 11, – 3)
Slope of the line m = tan θ = \(\frac{5}{12}\)
sin θ = \(\frac{5}{13}\), cos θ = \(\frac{12}{13}\)
The parametric equation of the line passing through the point (1, 2) making angle θ with x – axis is
Any point on this line is
(1 + r cos θ, 2 + r sin θ) ……… (1)
where r is the distance of any point from A (1, 2) on the line.
To find the point which is 13 units away from A (1, 2) on the line.
Substitute r = ± 13, cos θ = \(\frac{12}{13}\), sin θ = \(\frac{5}{13}\) in equation (1)
Required point = \(\left(1 \pm 13\left(\frac{12}{13}\right), 2 \pm 13\left(\frac{5}{13}\right)\right)\)
= (1 ± 12, 2 ± 5)
= (1 + 12, 2 + 5)
= (1 + 12, 2 + 5) or (1 – 12, 2 – 5)
= (13, 7) or (- 11, – 3)
Length of the train = 150 m
Constant velocity of the train = 12.5 m/s
(i) The equation of motion of the train:
Take time in seconds along the x-axis and distance in meters along the y-axis.
Let the train be at the origin.
∴ Length of the train = 150 m is the negative y-intercept
b = -150
The slope of the motion of the train m = 12.5 m/s
The equation of the line with slope-intercept form is
y = mx + b
∴ y = 12.5x – 150
which is the required equation of motion of the train.
(ii) Time taken to cross a pole:
The equation of motion is y = 12.5 x – 150
To find the time taken to cross the pole, Put y = 0
0 = 12.5 x – 150 ⇒ 12.5 x = 150
⇒ x = \(\frac{150}{12.5}\) = 12 sec
(iii) The time taken to cross the bridge of length 850 m
The equation of motion is y = 12.5 x – 150
To find the time taken to cross the bridge of length 850 m, put y = 850
850 = 12.5 x – 150
12.5 x = 850 + 150 = 1000
x = \(\frac{1000}{12.5}=\frac{10000}{125}\) = 80 sec
Length of the train = 150 m
Constant velocity of the train = 12.5 m/s
(i) The equation of motion of the train:
Take time in seconds along the x-axis and distance in meters along the y-axis.
Let the train be at the origin.
∴ Length of the train = 150 m is the negative y-intercept
b = -150
The slope of the motion of the train m = 12.5 m/s
The equation of the line with slope-intercept form is
y = mx + b
∴ y = 12.5x – 150
which is the required equation of motion of the train.
(ii) Time taken to cross a pole:
The equation of motion is y = 12.5 x – 150
To find the time taken to cross the pole, Put y = 0
0 = 12.5 x – 150 ⇒ 12.5 x = 150
⇒ x = \(\frac{150}{12.5}\) = 12 sec
(iii) The time taken to cross the bridge of length 850 m
The equation of motion is y = 12.5 x – 150
To find the time taken to cross the bridge of length 850 m, put y = 850
850 = 12.5 x – 150
12.5 x = 850 + 150 = 1000
x = \(\frac{1000}{12.5}=\frac{10000}{125}\) = 80 sec
- (a) Draw a graph showing the results.
- (b) Find the equation relating the length of the spring to the weight on it.
- (c) What is the actual length of the spring?
- (d) If the spring stretches to 9 cm long, how much weight should be added?
- (e) How long will the spring be when 6 kilograms of weight on it?
Choose the weight along the x-axis and Length along the y-axis.
(a)
(b) The points are(2, 3), (4, 4), (5, 4.5), (8, 6) The relation connecting weight and Length is the equation of the straight line joining the points (2, 3) and (4, 4)
x – 2 = 2(y – 3)
x – 2 = 2y – 6
x – 2y + 6 – 2 = 0
x – 2y + 4 = 0 —– (1)
which the required relation connecting weight and length.
(c) To find the actual length of the spring, put weight x = 0 in equation (1)
0 – 2y + 4 = 0 ⇒ 2y = 4 ⇒ y = 2
∴ The actual length of the spring is 2 cm.
(d) If the spring stretch to 9 cm long, To find the required weight, put y = 9, in equation (1)
(1) ⇒ x – 2 (9) + 4 = 0
x – 18 +4 = 0 ⇒ x = 14
Weight to be added is 14 kg.
(e) Next we find the length of the string when a weight of 6 kg is added.
Put x = 6 in equation (1)
6 – 2y + 4 = 0 ⇒ 2y = 10 ⇒ y = 5cm
∴ Required length is 5 cm.
Choose the weight along the x-axis and Length along the y-axis.
(a)
(b) The points are(2, 3), (4, 4), (5, 4.5), (8, 6) The relation connecting weight and Length is the equation of the straight line joining the points (2, 3) and (4, 4)
x – 2 = 2(y – 3)
x – 2 = 2y – 6
x – 2y + 6 – 2 = 0
x – 2y + 4 = 0 —– (1)
which the required relation connecting weight and length.
(c) To find the actual length of the spring, put weight x = 0 in equation (1)
0 – 2y + 4 = 0 ⇒ 2y = 4 ⇒ y = 2
∴ The actual length of the spring is 2 cm.
(d) If the spring stretch to 9 cm long, To find the required weight, put y = 9, in equation (1)
(1) ⇒ x – 2 (9) + 4 = 0
x – 18 +4 = 0 ⇒ x = 14
Weight to be added is 14 kg.
(e) Next we find the length of the string when a weight of 6 kg is added.
Put x = 6 in equation (1)
6 – 2y + 4 = 0 ⇒ 2y = 10 ⇒ y = 5cm
∴ Required length is 5 cm.
(i) Find the equation relating the quantity of gas in the cylinder to the days.
Given Total weight of cylinder = 29.5 kg
Weight of the gas inside the cylinder = 14.2 kg
Let x denote the number of days of consumption of the gas, y denote the quantity of gas inside a cylinder.
Initially x = 0 then y = 14.2
The corresponding point is (0, 14.2)
The gas inside the cylinder lasts for 24 days
∴ When x = 24, we have y = 0
The corresponding point is (24, 0)
∴ The linear relation between the quantity of gas in the cylinder to the number of days of consumption is the equation of the line joining the points (0, 14.2 ) and (24, 0).
which is the required relation
(ii) Draw the graph for the first 96 days:
The relation connecting the quantity of gas to the number of days of consumption is
y = –\(\frac{71}{120}\)x + 14.2
Let f(x) = –\(\frac{71}{120}\) x + 14.2
Here f(x) is a periodic function of period 24
∴ f(x + 24) = f(x)
When x = 0
f(0) = –\(\frac{71}{120}\) × 0 + 14.2 ⇒ y = 14.2
The corresponding point is (0, 14.2)
When x = 24
f(24) = –\(\frac{71}{120}\) × 0 + 14.2 ⇒ y = 14.2
⇒ f(24) = –\(\frac{71}{5}\) + 14.2
= – 14.2 + 14.2 = 0 ⇒ y = 0
Corresponding point is (24, 0)
When x = 48
f(48) = f(24 + 24 + 0) = f(24 + 0)
= f(o) = o
Corresponding point is (48, 0)
When x = 72
f(72) = f(24 + 24 + 24 + 0)
= f(24 + 24 + 0) = f(24 + 0)
= f(0) = 0
Corresponding point is (72, 0)
The required graph is
(i) Find the equation relating the quantity of gas in the cylinder to the days.
Given Total weight of cylinder = 29.5 kg
Weight of the gas inside the cylinder = 14.2 kg
Let x denote the number of days of consumption of the gas, y denote the quantity of gas inside a cylinder.
Initially x = 0 then y = 14.2
The corresponding point is (0, 14.2)
The gas inside the cylinder lasts for 24 days
∴ When x = 24, we have y = 0
The corresponding point is (24, 0)
∴ The linear relation between the quantity of gas in the cylinder to the number of days of consumption is the equation of the line joining the points (0, 14.2 ) and (24, 0).
which is the required relation
(ii) Draw the graph for the first 96 days:
The relation connecting the quantity of gas to the number of days of consumption is
y = –\(\frac{71}{120}\)x + 14.2
Let f(x) = –\(\frac{71}{120}\) x + 14.2
Here f(x) is a periodic function of period 24
∴ f(x + 24) = f(x)
When x = 0
f(0) = –\(\frac{71}{120}\) × 0 + 14.2 ⇒ y = 14.2
The corresponding point is (0, 14.2)
When x = 24
f(24) = –\(\frac{71}{120}\) × 0 + 14.2 ⇒ y = 14.2
⇒ f(24) = –\(\frac{71}{5}\) + 14.2
= – 14.2 + 14.2 = 0 ⇒ y = 0
Corresponding point is (24, 0)
When x = 48
f(48) = f(24 + 24 + 0) = f(24 + 0)
= f(o) = o
Corresponding point is (48, 0)
When x = 72
f(72) = f(24 + 24 + 24 + 0)
= f(24 + 24 + 0) = f(24 + 0)
= f(0) = 0
Corresponding point is (72, 0)
The required graph is
The equations of the given lines are
3x + 2y + 9 = 0 ——- (1)
12x + 8y – 15 = 0 ——- (2)
m 1 = m 2
∴ The given lines are parallel.
The equations of the given lines are
3x + 2y + 9 = 0 ——- (1)
12x + 8y – 15 = 0 ——- (2)
m 1 = m 2
∴ The given lines are parallel.
The equation of any line parallel to 5x – 4y + 3 = 0 is
5x – 4y + k = 0 ……….. (1)
The x – intercept of line (I) is obtained by putting
y = 0 in the equation.
(1) ⇒ 5x – 4(0) + k = 0
5x = – k ⇒ x = \(-\frac{k}{5}\)
Given that the x – intercept in 3
∴ \(-\frac{k}{5}\) = 3 ⇒ k = – 15
∴ The equation of the required line is
5x – 4y – 15 = 0
The equation of any line parallel to 5x – 4y + 3 = 0 is
5x – 4y + k = 0 ……….. (1)
The x – intercept of line (I) is obtained by putting
y = 0 in the equation.
(1) ⇒ 5x – 4(0) + k = 0
5x = – k ⇒ x = \(-\frac{k}{5}\)
Given that the x – intercept in 3
∴ \(-\frac{k}{5}\) = 3 ⇒ k = – 15
∴ The equation of the required line is
5x – 4y – 15 = 0
(i) The distance between the line ax + by + c = 0 and the point (x 1, y 1 ) is d = \(\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\)
Here (x 1, y 1 ) = (- 2, 4) and the equation of the line is 4x + 3y + 4 = 0
(ii) Given point (x 1, y 1 ) = (7, – 3)
Given line 4x + 3y + 4 = 0
(i) The distance between the line ax + by + c = 0 and the point (x 1, y 1 ) is d = \(\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\)
Here (x 1, y 1 ) = (- 2, 4) and the equation of the line is 4x + 3y + 4 = 0
(ii) Given point (x 1, y 1 ) = (7, – 3)
Given line 4x + 3y + 4 = 0
(i) Any line parallel to x + 3y – 4 = 0 will be of the form x + 3y + k = 0.
It passes through (1,-1) ⇒ 1 – 3 + k = 0 ⇒ k = 2
So the required line is x + 3y + 2 = 0
(ii) Any line perpendicular to 3x + 4y – 6 = 0 will be of the form 4x – 3y + k = 0.
It passes through (1, -1) ⇒ 4 + 3 + k = 0 ⇒ k = -7.
So the required line is 4x – 3y – 7 = 0
(i) Any line parallel to x + 3y – 4 = 0 will be of the form x + 3y + k = 0.
It passes through (1,-1) ⇒ 1 – 3 + k = 0 ⇒ k = 2
So the required line is x + 3y + 2 = 0
(ii) Any line perpendicular to 3x + 4y – 6 = 0 will be of the form 4x – 3y + k = 0.
It passes through (1, -1) ⇒ 4 + 3 + k = 0 ⇒ k = -7.
So the required line is 4x – 3y – 7 = 0
In a rhombus, the diagonal cut at right angles.
The given diagonal is 5x – y + 7 = 0 and (- 4, 7) is not a point on the diagonal.
So it will be a point on the other diagonal which is perpendicular to 5x – y + 7 = 0.
The equation of a line perpendicular to 5x – y + 7 = 0 will be of the form x + 5y + k = 0.
It passes through (-4, 7) ⇒ -4 + 5(7) + k = 0 ⇒ k = -31
So the equation of the other diagonal is x + 5y – 31 = 0
In a rhombus, the diagonal cut at right angles.
The given diagonal is 5x – y + 7 = 0 and (- 4, 7) is not a point on the diagonal.
So it will be a point on the other diagonal which is perpendicular to 5x – y + 7 = 0.
The equation of a line perpendicular to 5x – y + 7 = 0 will be of the form x + 5y + k = 0.
It passes through (-4, 7) ⇒ -4 + 5(7) + k = 0 ⇒ k = -31
So the equation of the other diagonal is x + 5y – 31 = 0
The equation of the straight line passing through the point of intersection of the lines.
4x – y + 3 = 0 and 5x + 2y + 7 = 0 is
( 4x – y + 3) + λ (5x + 2y + 7) = 0 ……… (1)
(i) Through the point (-1,2)
Given that line(1) passes through the point (-1, 2)
(1) ⇒ (4(-1) – 2 + 3) + λ (5(- 1) + 2(2) + 7) = 0
(-4 – 2 + 3) + λ (-5 + 4 + 7) = 0
– 3 + 6 λ = 0 ⇒ λ = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ The equation of the required line is
(4x – y + 3) + \(\frac{1}{2}\) (5x + 2y + 7) = 0
2(4x – y + 3) + (5x + 2y + 7) = 0
8x – 2y + 6 + 5x + 2y + 7 = 0
13x +13 = 0 ⇒ x + 1 = 0
(ii) Equation of a line parallel to x – y + 5 = 0 will be of the form x – y + k = 0.
It passes through (-1, -1) ⇒ -1 + 1 + k = 0 ⇒ k = 0.
So the required line is x – y = 0 ⇒ x = y.
(iii) Perpendicular to x – 2y + 1 = 0
Given that the line (1) perpendicular to the line
x – 2y + 1 = 0 ………….. (3)
(1) ⇒ (4x – y + 3) + λ (5x + 2y + 7) = 0
4x – y + 3 + 5λx + 2λy + 7λ = 0
(4 + 5λ)x + (2λ – 1 )y + (3 + 7λ) = 0 ……….. (4)
Slope of this line (3) = \(-\frac{4+5 \lambda}{2 \lambda-1}\)
Slope of line (2) = \(-\frac{1}{-2}=\frac{1}{2}\)
Given that line (3) and line (4) are perpendicular
4 + 5λ = 2(2λ – 1 )
4 + 5λ = 4λ – 2
λ = – 6
Substituting the value of λ in equation (1) we have
(4x – y + 3) – 6 (5x + 2y + 7) = 0
4x – y + 3 – 30x – 12y – 42 = 0
-26x – 13y – 39 =0
2x + y + 3 = 0
which is the required equation.
The equation of the straight line passing through the point of intersection of the lines.
4x – y + 3 = 0 and 5x + 2y + 7 = 0 is
( 4x – y + 3) + λ (5x + 2y + 7) = 0 ……… (1)
(i) Through the point (-1,2)
Given that line(1) passes through the point (-1, 2)
(1) ⇒ (4(-1) – 2 + 3) + λ (5(- 1) + 2(2) + 7) = 0
(-4 – 2 + 3) + λ (-5 + 4 + 7) = 0
– 3 + 6 λ = 0 ⇒ λ = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ The equation of the required line is
(4x – y + 3) + \(\frac{1}{2}\) (5x + 2y + 7) = 0
2(4x – y + 3) + (5x + 2y + 7) = 0
8x – 2y + 6 + 5x + 2y + 7 = 0
13x +13 = 0 ⇒ x + 1 = 0
(ii) Equation of a line parallel to x – y + 5 = 0 will be of the form x – y + k = 0.
It passes through (-1, -1) ⇒ -1 + 1 + k = 0 ⇒ k = 0.
So the required line is x – y = 0 ⇒ x = y.
(iii) Perpendicular to x – 2y + 1 = 0
Given that the line (1) perpendicular to the line
x – 2y + 1 = 0 ………….. (3)
(1) ⇒ (4x – y + 3) + λ (5x + 2y + 7) = 0
4x – y + 3 + 5λx + 2λy + 7λ = 0
(4 + 5λ)x + (2λ – 1 )y + (3 + 7λ) = 0 ……….. (4)
Slope of this line (3) = \(-\frac{4+5 \lambda}{2 \lambda-1}\)
Slope of line (2) = \(-\frac{1}{-2}=\frac{1}{2}\)
Given that line (3) and line (4) are perpendicular
4 + 5λ = 2(2λ – 1 )
4 + 5λ = 4λ – 2
λ = – 6
Substituting the value of λ in equation (1) we have
(4x – y + 3) – 6 (5x + 2y + 7) = 0
4x – y + 3 – 30x – 12y – 42 = 0
-26x – 13y – 39 =0
2x + y + 3 = 0
which is the required equation.
The equation of the given line is
12x + 5y + 2 = 0 ……… (1)
Equation of any line parallel to the line (1) is
12x + 5y + k = 0 ………. (2)
Given that line (2) is at a unit distance from the point (1, – 1)
k = – 7 ± 13
k = -7 + 13 or k = – 7 – 13
k = 6 or k = – 20
∴ The equation of the required lines are
12x + 5y + 6 = 0 and 12x + 5y – 20 = 0
The equation of the given line is
12x + 5y + 2 = 0 ……… (1)
Equation of any line parallel to the line (1) is
12x + 5y + k = 0 ………. (2)
Given that line (2) is at a unit distance from the point (1, – 1)
k = – 7 ± 13
k = -7 + 13 or k = – 7 – 13
k = 6 or k = – 20
∴ The equation of the required lines are
12x + 5y + 6 = 0 and 12x + 5y – 20 = 0
The equation of the given line is
3x + 4y – 6 = 0 …………. (1)
The equation of any line perpendicular to line (1) is
4x – 3y + k = 0 …………. (2)
Given that this line is 4 units from the point (2, 1)
± 4 = \(\frac{5+k}{5}\)
5 + k = ± 20
k = ± 20 – 5
k = 20 – 5 or k = -20
k = 15 or k = -25
∴ The equation of the required lines are
4x – 3y + 15 = 0 and 4x – 3y – 25 = 0
The equation of the given line is
3x + 4y – 6 = 0 …………. (1)
The equation of any line perpendicular to line (1) is
4x – 3y + k = 0 …………. (2)
Given that this line is 4 units from the point (2, 1)
± 4 = \(\frac{5+k}{5}\)
5 + k = ± 20
k = ± 20 – 5
k = 20 – 5 or k = -20
k = 15 or k = -25
∴ The equation of the required lines are
4x – 3y + 15 = 0 and 4x – 3y – 25 = 0
The equation of the given line is
2x + 3y = 10 ………….. (1)
The equation of any line parallel to (1) is
2x + 3y = k …………. (2)
Given that the sum of the intercepts of the line (2) on the axes is 15
∴ The equation of the required line is 2x + 3y = 18
The equation of the given line is
2x + 3y = 10 ………….. (1)
The equation of any line parallel to (1) is
2x + 3y = k …………. (2)
Given that the sum of the intercepts of the line (2) on the axes is 15
∴ The equation of the required line is 2x + 3y = 18
The coordinate of the foot of the perpendicular from the point (x 1, y 1 ) on the line ax + by + c = 0 is
∴ The coordinate of the foot of the perpendicular from the point (- 10, – 2) on the line x + y – 2 = 0 is
x + 10 = y + 2 = \(\frac{14}{2}\)
x + 10 = y + 2 = 7
x + 10 = 7, y + 2 = 7
x = – 3, y = 5
∴ The required foot of the perpendicular is (- 3, 5).
Length of the perpendicular
The coordinate of the foot of the perpendicular from the point (x 1, y 1 ) on the line ax + by + c = 0 is
∴ The coordinate of the foot of the perpendicular from the point (- 10, – 2) on the line x + y – 2 = 0 is
x + 10 = y + 2 = \(\frac{14}{2}\)
x + 10 = y + 2 = 7
x + 10 = 7, y + 2 = 7
x = – 3, y = 5
∴ The required foot of the perpendicular is (- 3, 5).
Length of the perpendicular
Given P 1 is the length of the perpendicular from the origin to the straight line
x sec θ + y cosec θ – 2a = 0
Also given P 2 is the length of the perpendicular from the origin to the straight line
x cos θ – y sin θ – a cos 2θ = 0
Given P 1 is the length of the perpendicular from the origin to the straight line
x sec θ + y cosec θ – 2a = 0
Also given P 2 is the length of the perpendicular from the origin to the straight line
x cos θ – y sin θ – a cos 2θ = 0
(i) 12x + 5y = 7 and 12x + 5y + 7 = 0
The equation of the given lines are
12x + 5y – 7 = 0 ……….. (1)
12x + 5y + 7 = 0 ……….. (2)
The distance between the parallel lines
ax + by + c 1 = 0 and ax + by + c 2 = 0 is
The equation of any line parallel to (1) is
The distance cannot be negative
∴ Required distance = \(\frac{14}{13}\)
(ii) 3x – 4y + 5 = 0 and 6x – 8y -15 = 0
The equation of the given lines are
3x – 4y + 5 = 0 ……….. (1)
6x – 8y – 15 = 0
3x – 4y – 1 = 0 ………… (2)
The distance between the parallel lines (1) and (2) is
(i) 12x + 5y = 7 and 12x + 5y + 7 = 0
The equation of the given lines are
12x + 5y – 7 = 0 ……….. (1)
12x + 5y + 7 = 0 ……….. (2)
The distance between the parallel lines
ax + by + c 1 = 0 and ax + by + c 2 = 0 is
The equation of any line parallel to (1) is
The distance cannot be negative
∴ Required distance = \(\frac{14}{13}\)
(ii) 3x – 4y + 5 = 0 and 6x – 8y -15 = 0
The equation of the given lines are
3x – 4y + 5 = 0 ……….. (1)
6x – 8y – 15 = 0
3x – 4y – 1 = 0 ………… (2)
The distance between the parallel lines (1) and (2) is
(i) Equation of lines perpendicular to 3x + 4y – 12 = 0 will be of the form 4x – 3y + k = 0, k ∈ R
(ii) Equation of lines parallel to 3x + 4y – 12 = 0 will be of the form 3x + 4y + k = 0, k ∈ R
(i) Equation of lines perpendicular to 3x + 4y – 12 = 0 will be of the form 4x – 3y + k = 0, k ∈ R
(ii) Equation of lines parallel to 3x + 4y – 12 = 0 will be of the form 3x + 4y + k = 0, k ∈ R
Slope of the line AB
m = tan θ = \(\frac{1-0}{3-2}\)
tan θ = 1
θ = 45°
∴ The line AB makes an angle 45° with x-axis.
Given that the line AB is rotated through an angle of 15° about the point A in the anticlockwise direction.
∴ The angle made by the new line AB’ is 45° + 15° = 60°
Slope of the new line AB’ is m 1 = tan 60° = √3
∴ The equation of the new line AB’ is the equation of the straight line passing through the point A (2, 0) and having slope m 1 = √3
y – 0 = √3 (x – 2)
y = √3x – 2√3
√3x – y – 2√3 = 0
Slope of the line AB
m = tan θ = \(\frac{1-0}{3-2}\)
tan θ = 1
θ = 45°
∴ The line AB makes an angle 45° with x-axis.
Given that the line AB is rotated through an angle of 15° about the point A in the anticlockwise direction.
∴ The angle made by the new line AB’ is 45° + 15° = 60°
Slope of the new line AB’ is m 1 = tan 60° = √3
∴ The equation of the new line AB’ is the equation of the straight line passing through the point A (2, 0) and having slope m 1 = √3
y – 0 = √3 (x – 2)
y = √3x – 2√3
√3x – y – 2√3 = 0
Let P(1, 2) and (5, 3) are the given points.
By the property of reflection,
∠XAB = ∠OAP = θ
(Angle of incidence = Angle of reflection)
Slope of the line OA (x – axis) m 1 = 0
Slope of the line joining the points P (1, 2) and A (x, 0)
Slope of AP, m 2 = \(\frac{2-0}{1-x}=\frac{2}{1-x}\)
Slope of the line joining the points B (5, 3 ) and A (x, 0)
Slope of AP, m 3 = \(\frac{3-0}{5-x}=\frac{3}{5-x}\)
From equations (1) and (2)
\(\frac{3}{5-x}\) = –\(\frac{2}{1-x}\)
3(1 – x) = – 2 (5 – x)
3 – 3x = – 10 + 2x
2x + 3x = 10 + 3
5x = 13 ⇒ x = \(\frac{13}{5}\)
∴ The required point A is \(\left(\frac{13}{5}, 0\right)\)
Let P(1, 2) and (5, 3) are the given points.
By the property of reflection,
∠XAB = ∠OAP = θ
(Angle of incidence = Angle of reflection)
Slope of the line OA (x – axis) m 1 = 0
Slope of the line joining the points P (1, 2) and A (x, 0)
Slope of AP, m 2 = \(\frac{2-0}{1-x}=\frac{2}{1-x}\)
Slope of the line joining the points B (5, 3 ) and A (x, 0)
Slope of AP, m 3 = \(\frac{3-0}{5-x}=\frac{3}{5-x}\)
From equations (1) and (2)
\(\frac{3}{5-x}\) = –\(\frac{2}{1-x}\)
3(1 – x) = – 2 (5 – x)
3 – 3x = – 10 + 2x
2x + 3x = 10 + 3
5x = 13 ⇒ x = \(\frac{13}{5}\)
∴ The required point A is \(\left(\frac{13}{5}, 0\right)\)
Let the given line be PQ whose equation is
5x = y + 7
5x – y – 7 = 0 ——- (1)
Let AB be the line perpendicular to the line PQ such that the area of the triangle OAB is 10 units.
The equation of the line AB is
– x – 5y + k = 0
x + 5y – k = 0
x + 5y = k ——- (2)
∴ A is (k, 0) and B is (0, \(\frac{\mathrm{k}}{5}\))
OA = k and OB = \(\frac{\mathrm{k}}{5}\)
Area of ∆OAB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × k × \(\frac{\mathrm{k}}{5}\)
Given area of ∆ OAB = 10
∴ \(\frac{\mathrm{k}^{2}}{10}\) = 10
k 2 = 100 ⇒ k = ±10
∴ The required equation of the straight line is
x + 5y = ±10
Let the given line be PQ whose equation is
5x = y + 7
5x – y – 7 = 0 ——- (1)
Let AB be the line perpendicular to the line PQ such that the area of the triangle OAB is 10 units.
The equation of the line AB is
– x – 5y + k = 0
x + 5y – k = 0
x + 5y = k ——- (2)
∴ A is (k, 0) and B is (0, \(\frac{\mathrm{k}}{5}\))
OA = k and OB = \(\frac{\mathrm{k}}{5}\)
Area of ∆OAB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × k × \(\frac{\mathrm{k}}{5}\)
Given area of ∆ OAB = 10
∴ \(\frac{\mathrm{k}^{2}}{10}\) = 10
k 2 = 100 ⇒ k = ±10
∴ The required equation of the straight line is
x + 5y = ±10
The image of the point (x 1, y 1 ) about the line ax + by + c = 0 is
∴ The image of the point (- 2, 3) about the line x + 2y – 9 = 0 is
∴ The required point is (0, 7)
The image of the point (x 1, y 1 ) about the line ax + by + c = 0 is
∴ The image of the point (- 2, 3) about the line x + 2y – 9 = 0 is
∴ The required point is (0, 7)
(i) Draw graph of the cost as x goes from 0 to 50 copies:
Let x represent the number of copies and y represent the cost of photocopying.
Given photo copying charge for 1 copy is Rs. 1.50 for the first 10 copies and Rs. 1.00 per copy after the 10 th copy.
∴ The relation connecting the number of copies and cost of photocopying charge is given by
y = 1.50x, 0 ≤ x ≤ 10
y = 10(1.50) + (x – 10) (1)
y = 15 + x – 10
y = x + 5 ………… (1)
The graph for 0 to 50 copies:
When x = 10, y = 1.50 × x
⇒ y = 1.50 × 10 = 15
The corresponding point is (10, 15)
When x = 20, y = x + 5
⇒ y = 20 + 5 = 25
The corresponding point is (20, 25)
When x = 30, y = x + 5
⇒ y = 30 + 5 = 35
The corresponding point is (30, 35)
When x = 40, y = 40 + 5 = 45
The corresponding point is (40, 45 )
When x = 50, y = 50 + 5 = 55
The corresponding point is (50, 55 )
The cost of 40 copies is the value of y
When x = 40, y = 40 + 5 = 45 rupees
Cost of 40 copies = 45 rupees
(i) Draw graph of the cost as x goes from 0 to 50 copies:
Let x represent the number of copies and y represent the cost of photocopying.
Given photo copying charge for 1 copy is Rs. 1.50 for the first 10 copies and Rs. 1.00 per copy after the 10 th copy.
∴ The relation connecting the number of copies and cost of photocopying charge is given by
y = 1.50x, 0 ≤ x ≤ 10
y = 10(1.50) + (x – 10) (1)
y = 15 + x – 10
y = x + 5 ………… (1)
The graph for 0 to 50 copies:
When x = 10, y = 1.50 × x
⇒ y = 1.50 × 10 = 15
The corresponding point is (10, 15)
When x = 20, y = x + 5
⇒ y = 20 + 5 = 25
The corresponding point is (20, 25)
When x = 30, y = x + 5
⇒ y = 30 + 5 = 35
The corresponding point is (30, 35)
When x = 40, y = 40 + 5 = 45
The corresponding point is (40, 45 )
When x = 50, y = 50 + 5 = 55
The corresponding point is (50, 55 )
The cost of 40 copies is the value of y
When x = 40, y = 40 + 5 = 45 rupees
Cost of 40 copies = 45 rupees
The equations of the given straight lines are
y = 5x + b ……….. (1)
3x – 4y = 6 ……….. (2)
To find atleast two equations from the family y = 5x + b for which b is an integer and x – coordinate of the point of intersection of (1) and (2) is an integer. Solving (1)and (2) using equation (1) inequation (2) (2) ⇒ 3x – 4 (5x + b) = 6
3x – 20x – 4b = 6
-17x = 6 + 4b
The corresponding equation of the line is = 5x + 7
When b = – 10, we have x = \(\frac{6-40}{-17}\)
= \(\frac{-34}{-17}\) = 2
The corresponding equation of the line is y = 5x – 10
Thus y = 5x + 7 and y = 5x – 10 are the two straight lines belonging to the family such that b is an integer and the x – coordinate of the point of intersection with the line (2) is an integer.
The equations of the given straight lines are
y = 5x + b ……….. (1)
3x – 4y = 6 ……….. (2)
To find atleast two equations from the family y = 5x + b for which b is an integer and x – coordinate of the point of intersection of (1) and (2) is an integer. Solving (1)and (2) using equation (1) inequation (2) (2) ⇒ 3x – 4 (5x + b) = 6
3x – 20x – 4b = 6
-17x = 6 + 4b
The corresponding equation of the line is = 5x + 7
When b = – 10, we have x = \(\frac{6-40}{-17}\)
= \(\frac{-34}{-17}\) = 2
The corresponding equation of the line is y = 5x – 10
Thus y = 5x + 7 and y = 5x – 10 are the two straight lines belonging to the family such that b is an integer and the x – coordinate of the point of intersection with the line (2) is an integer.
The equations of the given lines are
y = mx – 3 ………. (1)
x – y = 6 ………. (2)
Solving equations (1) and (2)
(2) ⇒ x – (mx – 3 ) = 6
x – mx + 3 = 6
x (1 – m) = 3
x = \(\frac{3}{1-\mathrm{m}}\) …….. (3)
From equation (3) let us find the values of x and m for which they are integers. The only values of m for which, x is an integer are m = 0, 2, -2
When m = 0, x = \(\frac{3}{1-0}\) = 3
The corresponding equation is
y = 0. x – 3
y + 3 = 0
When m = 2, x = \(\frac{3}{1-2}\) = \(\frac{3}{-1}\) = – 3
The corresponding equation is y = -2x + 3
2x + y – 3 = 0
When m = – 2, x = \(\frac{3}{1+2}\) = \(\frac{3}{3}\) = 1
The corresponding equation is
y = – 2 x + 3
2x + y – 3 = 0
∴ The required equations of the lines are
y + 3 = 0, 2x – y – 3 = 0 and
2x + y – 3 = 0
The equations of the given lines are
y = mx – 3 ………. (1)
x – y = 6 ………. (2)
Solving equations (1) and (2)
(2) ⇒ x – (mx – 3 ) = 6
x – mx + 3 = 6
x (1 – m) = 3
x = \(\frac{3}{1-\mathrm{m}}\) …….. (3)
From equation (3) let us find the values of x and m for which they are integers. The only values of m for which, x is an integer are m = 0, 2, -2
When m = 0, x = \(\frac{3}{1-0}\) = 3
The corresponding equation is
y = 0. x – 3
y + 3 = 0
When m = 2, x = \(\frac{3}{1-2}\) = \(\frac{3}{-1}\) = – 3
The corresponding equation is y = -2x + 3
2x + y – 3 = 0
When m = – 2, x = \(\frac{3}{1+2}\) = \(\frac{3}{3}\) = 1
The corresponding equation is
y = – 2 x + 3
2x + y – 3 = 0
∴ The required equations of the lines are
y + 3 = 0, 2x – y – 3 = 0 and
2x + y – 3 = 0
Separate equations are x – 2y – 3 = 0; x + y + 5 = 0
So the combined equation is (x – 2y – 3) (x + y + 5) = 0
x 2 + xy + 5x – 2y 2 – 2xy – 10y – 3x – 3y – 15 = 0
(i.e) x 2 – 2y 2 – xy + 2x – 13y – 15 = 0
Separate equations are x – 2y – 3 = 0; x + y + 5 = 0
So the combined equation is (x – 2y – 3) (x + y + 5) = 0
x 2 + xy + 5x – 2y 2 – 2xy – 10y – 3x – 3y – 15 = 0
(i.e) x 2 – 2y 2 – xy + 2x – 13y – 15 = 0
The equation of the given pair of straight lines is
4x 2 + 4xy + y 2 – 6x – 3y – 4 = 0 ………. (1)
Compare this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 ………. (2)
a = 4, 2h = 4, b = 1, 2g = -6,
2f = -3, c = – 4
The condition for parallelism is
h 2 – ab = 0
2 2 – (4) (1) = 4 – 4 = 0
∴ The given pair of straight lines represents a pair of parallel straight lines.
The equation of the given pair of straight lines is
4x 2 + 4xy + y 2 – 6x – 3y – 4 = 0 ………. (1)
Compare this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 ………. (2)
a = 4, 2h = 4, b = 1, 2g = -6,
2f = -3, c = – 4
The condition for parallelism is
h 2 – ab = 0
2 2 – (4) (1) = 4 – 4 = 0
∴ The given pair of straight lines represents a pair of parallel straight lines.
Comparing the given equation with the general form a = 2,h = 3/2, b = -2,g= 3/2, f = 1/2 and c = 1
Condition for two lines to be perpendicular is a + b = 0. Here a + b = 2 – 2 = 0
⇒ The given equation represents a pair of perpendicular lines.
Comparing the given equation with the general form a = 2,h = 3/2, b = -2,g= 3/2, f = 1/2 and c = 1
Condition for two lines to be perpendicular is a + b = 0. Here a + b = 2 – 2 = 0
⇒ The given equation represents a pair of perpendicular lines.
The equation of the given pair of lines is
2x 2 – xy – 3y 2 – 6x + 19y – 20 = 0 —- (1)
Compare this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 —- (2)
a = 2, 2h = – 1, b = – 3, 2g = – 6, 2f = 19, c = – 20
h 2 – ab = \(\left(-\frac{1}{2}\right)^{2}\) – (2)(-3)
= \(\frac{1}{4}\) + 6 ≠ 0
∴ The given line (1) is not parallel.
∴ They are intersecting lines.
Let θ be the angle between the lines.
Taking the acute angle θ = tan -1 (5)
The equation of the given pair of lines is
2x 2 – xy – 3y 2 – 6x + 19y – 20 = 0 —- (1)
Compare this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 —- (2)
a = 2, 2h = – 1, b = – 3, 2g = – 6, 2f = 19, c = – 20
h 2 – ab = \(\left(-\frac{1}{2}\right)^{2}\) – (2)(-3)
= \(\frac{1}{4}\) + 6 ≠ 0
∴ The given line (1) is not parallel.
∴ They are intersecting lines.
Let θ be the angle between the lines.
Taking the acute angle θ = tan -1 (5)
Let OP be the given line y = x having slope
m = 1 = tan 45°
Given that OA is the line making angle α with the line y = x.
Slope of the line OA = tan (45° – α)
The equation of OA is the equation of the line passing through the point (0, 0) having slope tan (45° – α).
y – 0 = tan(45°- α) (x – 0)
y = x tan (45°- α)
x tan (45° – α) – y = 0
Also given the line OB makes an angle α with the line y = x.
Slope of the line OB = tan(45° + α)
The equation of OB is the equation of the line passing through the point (0, 0) having slope tan(45° + α)
y – 0 = tan (45° + α) (x – 0)
y = x tan (45° + α)
x tan (45° + α) – y = 0 ………. (2)
The combined equation is
(x tan(45°- α) – y) (x tan(45°+ α) – y) = 0
x 2 tan(45° – α) tan(45° + α) – xy tan (45° – α) – xy tan(45° + α) + yz = 0
Let OP be the given line y = x having slope
m = 1 = tan 45°
Given that OA is the line making angle α with the line y = x.
Slope of the line OA = tan (45° – α)
The equation of OA is the equation of the line passing through the point (0, 0) having slope tan (45° – α).
y – 0 = tan(45°- α) (x – 0)
y = x tan (45°- α)
x tan (45° – α) – y = 0
Also given the line OB makes an angle α with the line y = x.
Slope of the line OB = tan(45° + α)
The equation of OB is the equation of the line passing through the point (0, 0) having slope tan(45° + α)
y – 0 = tan (45° + α) (x – 0)
y = x tan (45° + α)
x tan (45° + α) – y = 0 ………. (2)
The combined equation is
(x tan(45°- α) – y) (x tan(45°+ α) – y) = 0
x 2 tan(45° – α) tan(45° + α) – xy tan (45° – α) – xy tan(45° + α) + yz = 0
Equation of a line perpendicular to 2x – 3y + 1 = 0 is of the form 3x + 2y + k = 0.
It passes through (1, 3) ⇒ 3 + 6 + k = 0 ⇒ k = – 9
So the line is 3x + 2y – 9 = 0
The equation of a line perpendicular to 5x + y – 3 = 0 will be of the form x – 5y + k = 0.
It passes through (1, 3) ⇒ 1 – 15 + k = 0 ⇒ k = 14
So the line is x – 5y + 14 = 0.
The equation of the lines is 3x + 2y – 9 = 0 and x – 5y + 14 = 0
Their combined equation is (3x + 2y – 9)(x – 5y + 14) = 0
(i.e) 3x 2 – 15xy + 42x + 2xy – 10y 2 + 28y – 9x + 45y – 126 = 0
(i.e) 3x 2 – 13xy – 10y 2 + 33x + 73y – 126 = 0
Equation of a line perpendicular to 2x – 3y + 1 = 0 is of the form 3x + 2y + k = 0.
It passes through (1, 3) ⇒ 3 + 6 + k = 0 ⇒ k = – 9
So the line is 3x + 2y – 9 = 0
The equation of a line perpendicular to 5x + y – 3 = 0 will be of the form x – 5y + k = 0.
It passes through (1, 3) ⇒ 1 – 15 + k = 0 ⇒ k = 14
So the line is x – 5y + 14 = 0.
The equation of the lines is 3x + 2y – 9 = 0 and x – 5y + 14 = 0
Their combined equation is (3x + 2y – 9)(x – 5y + 14) = 0
(i.e) 3x 2 – 15xy + 42x + 2xy – 10y 2 + 28y – 9x + 45y – 126 = 0
(i.e) 3x 2 – 13xy – 10y 2 + 33x + 73y – 126 = 0
(i) Factorising 3x 2 + 2xy – y 2 we get
3x 2 + 3xy – xy – y 2 = 3x (x + y) – y (x + y)
= (3 x – y)(x + y)
So 3x 2 + 2xy – y 2 = 0 ⇒ (3x – y) (x + y) = 0
⇒ 3x – y = 0 and x + y = 0
(ii) 6(x – 1) 2 + 5 ( x – 1) (y – 2) – 4(y – 3 ) 2 = 0
Let X = x – 1 and Y = y – 2
∴ The given equation becomes
6X 2 + 5XY – 4Y 2 = 0
6X 2 + 8 XY – 3XY – 4Y 2 = 0
2X(3X + 4Y) – Y (3X + 4Y) = 0
(2X – Y) (3X + 4Y) = 0
2X – Y = 0 and 3X + 4Y = 0
Substituting for X and Y, we have
2X – Y = 0 ⇒ 2(x – 1) – (y – 2) = 0
⇒ 2x – 2 – y + 2 = 0
⇒ 2x – y = 0
3X + 4Y = 0 ⇒ 3 (x – 1) + 4 ( y – 2 ) = 0
⇒ 3x – 3 + 4y – 8 = 0
⇒ 3x + 4y – 11 = 0
∴ The separate equations are
2x – y = 0 and 3x + 4y – 11 = 0
(iii) 2x 2 – xy – 3y 2 – 6x + 19y – 20 = 0
The given equation is
2x 2 – xy – 3y 2 – 6x + 19y – 20 = 0 ……… (1)
2x 2 – xy – 3y 2 = 2x 2 – 3xy + 2xy – 3y 2
= x (2x – 3y) + y (2x – 3y )
= (x + y) (2x – 3y)
Let the separate equation of the straight lines be
x + y + 1 = 0 and 2x – 3y + m = 0 …….. (A)
(1) ⇒ 2x 2 – xy – 3y 2 – 6x + 19y – 20 = (x + y + 1) (2x – 3y + m)
Equating the coeffident of x, y and constant term on both sides, we have
-6 = m + 2l ………. (2)
19 = m – 3l ………. (3)
lm = – 20 ………. (4)
Solving equations (2), (3) and (4) we have
Substituting the values of l and m in equation (A) the required separate equations of the lines are
x + y – 5 = 0 and 2x – 3y + 4 = 0
(i) Factorising 3x 2 + 2xy – y 2 we get
3x 2 + 3xy – xy – y 2 = 3x (x + y) – y (x + y)
= (3 x – y)(x + y)
So 3x 2 + 2xy – y 2 = 0 ⇒ (3x – y) (x + y) = 0
⇒ 3x – y = 0 and x + y = 0
(ii) 6(x – 1) 2 + 5 ( x – 1) (y – 2) – 4(y – 3 ) 2 = 0
Let X = x – 1 and Y = y – 2
∴ The given equation becomes
6X 2 + 5XY – 4Y 2 = 0
6X 2 + 8 XY – 3XY – 4Y 2 = 0
2X(3X + 4Y) – Y (3X + 4Y) = 0
(2X – Y) (3X + 4Y) = 0
2X – Y = 0 and 3X + 4Y = 0
Substituting for X and Y, we have
2X – Y = 0 ⇒ 2(x – 1) – (y – 2) = 0
⇒ 2x – 2 – y + 2 = 0
⇒ 2x – y = 0
3X + 4Y = 0 ⇒ 3 (x – 1) + 4 ( y – 2 ) = 0
⇒ 3x – 3 + 4y – 8 = 0
⇒ 3x + 4y – 11 = 0
∴ The separate equations are
2x – y = 0 and 3x + 4y – 11 = 0
(iii) 2x 2 – xy – 3y 2 – 6x + 19y – 20 = 0
The given equation is
2x 2 – xy – 3y 2 – 6x + 19y – 20 = 0 ……… (1)
2x 2 – xy – 3y 2 = 2x 2 – 3xy + 2xy – 3y 2
= x (2x – 3y) + y (2x – 3y )
= (x + y) (2x – 3y)
Let the separate equation of the straight lines be
x + y + 1 = 0 and 2x – 3y + m = 0 …….. (A)
(1) ⇒ 2x 2 – xy – 3y 2 – 6x + 19y – 20 = (x + y + 1) (2x – 3y + m)
Equating the coeffident of x, y and constant term on both sides, we have
-6 = m + 2l ………. (2)
19 = m – 3l ………. (3)
lm = – 20 ………. (4)
Solving equations (2), (3) and (4) we have
Substituting the values of l and m in equation (A) the required separate equations of the lines are
x + y – 5 = 0 and 2x – 3y + 4 = 0
The equation of the given straight line is
ax 2 + 2hxy + by 2 = 0 ………… (1)
Given that the slopes of the straight lines are m and 2m
The equation of the given straight line is
ax 2 + 2hxy + by 2 = 0 ………… (1)
Given that the slopes of the straight lines are m and 2m
The equation of the given straight line is
ax 2 + 2hxy + by 2 = 0 ……….. (1)
Given that the slopes of the lines are m and 3m.
The equation of the given straight line is
ax 2 + 2hxy + by 2 = 0 ……….. (1)
Given that the slopes of the lines are m and 3m.
The equation of the given pair of lines is
x 2 – 4xy + y 2 = 0 ……….. (1)
The equation of the line PQ is
x + y – 2 = 0
y = 2 – x ……….. (2)
To find the coordinates of P and Q,.
Solve equations (1) and (2)
(1) ⇒ x 2 – 4x ( 2 – x) + ( 2 – x) 2 = 0
x 2 – 8x + 4x 2 + 4 – 4x + x 2 = 0
6x 2 – 12x + 4 = 0
3x 2 – 6x + 2 = 0
The midpoint of PQ is
The equation of the median drawn from 0 is the equation of the line joining 0 (0, 0) and D (1, 1)
∴ The required equation is x = y
The equation of the given pair of lines is
x 2 – 4xy + y 2 = 0 ……….. (1)
The equation of the line PQ is
x + y – 2 = 0
y = 2 – x ……….. (2)
To find the coordinates of P and Q,.
Solve equations (1) and (2)
(1) ⇒ x 2 – 4x ( 2 – x) + ( 2 – x) 2 = 0
x 2 – 8x + 4x 2 + 4 – 4x + x 2 = 0
6x 2 – 12x + 4 = 0
3x 2 – 6x + 2 = 0
The midpoint of PQ is
The equation of the median drawn from 0 is the equation of the line joining 0 (0, 0) and D (1, 1)
∴ The required equation is x = y
The equation of the given pair of straight lines is
6x 2 + 5xy – py 2 + 7x + qy – 5 = 0 ……….. (1)
Given that equation (1) represents a pair of perpendicular straight lines.
∴ Coefficient of x 2 + coefficient of y 2 = 0
6 – p = 0 ⇒ p = 6
6x 2 + 5xy – 6y 2 = 6x 2 + 9xy – 4xy – 6y 2
= 3x(2x + 3y) – 2y (2x + 3y)
= (2x + 3y) (3x – 2y)
Let the separate equation of the straight lines be
2x + 3y + 1 = 0 and 3x – 2y + m = 0
6x 2 + 5xy – 6y 2 + 7x + qy – 5
= (2x + 3y + 1)(3x – 2y + m)
Comparing the coefficients of x, y and constant terms on both sides
2m + 3l = 7 ………… (2)
3m – 2l = q ……….. (3)
lm = – 5 …….. (4)
Equation (4) ⇒ l = 1, m = – 5
or l = – 1, m = 5
When l = 1, m = – 5, equation (2) does not satisfy.
∴ l = – 1, m = 5
Substituting in equation (3)
3 (5) – 2(-1) = q ⇒ q = 17
∴ The required values are p = 6, q = 17
The equation of the given pair of straight lines is
6x 2 + 5xy – py 2 + 7x + qy – 5 = 0 ……….. (1)
Given that equation (1) represents a pair of perpendicular straight lines.
∴ Coefficient of x 2 + coefficient of y 2 = 0
6 – p = 0 ⇒ p = 6
6x 2 + 5xy – 6y 2 = 6x 2 + 9xy – 4xy – 6y 2
= 3x(2x + 3y) – 2y (2x + 3y)
= (2x + 3y) (3x – 2y)
Let the separate equation of the straight lines be
2x + 3y + 1 = 0 and 3x – 2y + m = 0
6x 2 + 5xy – 6y 2 + 7x + qy – 5
= (2x + 3y + 1)(3x – 2y + m)
Comparing the coefficients of x, y and constant terms on both sides
2m + 3l = 7 ………… (2)
3m – 2l = q ……….. (3)
lm = – 5 …….. (4)
Equation (4) ⇒ l = 1, m = – 5
or l = – 1, m = 5
When l = 1, m = – 5, equation (2) does not satisfy.
∴ l = – 1, m = 5
Substituting in equation (3)
3 (5) – 2(-1) = q ⇒ q = 17
∴ The required values are p = 6, q = 17
The equation of the given pair of straight lines is
12x 2 + 7xy – 12y 2 – x + 7y + k = 0 ………. (1)
Compare this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2f y + c = 0 ……….. (2)
a = 12, 2h = 7, b = – 12,
2g = – 1, 2f = 7, k = c
a = 12, h = \(\frac{7}{2}\), b = – 12,
g = \(-\frac{1}{2}\), f = \(\frac{7}{2}\), c = k
The condition for a second degree equation in x and y to represent a pair of straight lines is
abc + 2fgh – af 2 – bg 2 – ch 2 = 0
Substituting the values
Coefficient of x 2 + coefficient of y 2 = 12 – 12 = 0
∴ The given pair of straight lines are perpendicular and hence they are intersecting lines.
The equation of the given pair of straight lines is
12x 2 + 7xy – 12y 2 – x + 7y + k = 0 ………. (1)
Compare this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2f y + c = 0 ……….. (2)
a = 12, 2h = 7, b = – 12,
2g = – 1, 2f = 7, k = c
a = 12, h = \(\frac{7}{2}\), b = – 12,
g = \(-\frac{1}{2}\), f = \(\frac{7}{2}\), c = k
The condition for a second degree equation in x and y to represent a pair of straight lines is
abc + 2fgh – af 2 – bg 2 – ch 2 = 0
Substituting the values
Coefficient of x 2 + coefficient of y 2 = 12 – 12 = 0
∴ The given pair of straight lines are perpendicular and hence they are intersecting lines.
The given equation of the pair of straight line is
12x 2 + 2kxy + 2y 2 + 11x – 5y + 2 = 0 ……… (1)
Compare this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 ………. (2)
a = 12, 2h = 2k, b = 2,
2g = 11, 2f = – 5, c = 2,
a =12, h = k, b = 2,
g = \(\frac{11}{2}\), f = –\(\frac{5}{2}\), c = 2,
The condition for a second degree equation in x and y to represent a pair of straight lines is
abc + 2fgh – af 2 – bg 2 – ch 2 = 0
96 – 55k – 150 – 121 – 4k 2 = o
– 4k 2 – 55k – 175 = 0
4k 2 + 55k + 175 = 0
∴ The given equation represents a pair of straight lines when
k = – 5 and k = \(\frac{-35}{4}\)
The given equation of the pair of straight line is
12x 2 + 2kxy + 2y 2 + 11x – 5y + 2 = 0 ……… (1)
Compare this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 ………. (2)
a = 12, 2h = 2k, b = 2,
2g = 11, 2f = – 5, c = 2,
a =12, h = k, b = 2,
g = \(\frac{11}{2}\), f = –\(\frac{5}{2}\), c = 2,
The condition for a second degree equation in x and y to represent a pair of straight lines is
abc + 2fgh – af 2 – bg 2 – ch 2 = 0
96 – 55k – 150 – 121 – 4k 2 = o
– 4k 2 – 55k – 175 = 0
4k 2 + 55k + 175 = 0
∴ The given equation represents a pair of straight lines when
k = – 5 and k = \(\frac{-35}{4}\)
The given equation of the pair of straight line is
9x 2 – 24xy + by 2 – 12x + 16y – 12 = 0 ………… (1)
9x 2 – 24xy + 16y 2 = 9x 2 – 12xy – 12xy + 16y 2
= 3x (3x – 4y) – 4y (3x – 4y)
= (3x – 4y) (3x – 4y)
Let the separate equation of the straight lines be
3x – 4y + 1 = 0 and 3x – 4y + m = 0
9x 2 – 24xy + 16y 2 – 12x + 16y – 12
= (3x – 4y + l) ( 3x – 4y + m )
Comparing the coefficients of x, y and constant terms on both sides
3l + 3m = – 12
l + m = – 4 ……….. (2)
– 4l – 4m = 16
l + m = – 4 ………… (3)
lm = – 12 ……….. (4)
(l – m) 2 = (l + m) 2 – 4lm
= (- 4) 2 – 4 × – 12
= 16 + 48 = 64
l – m = √64 = 8
l – m = 8 ………… (5)
Solving equations (2) and (5 ), we have
(2) ⇒ 2 + m = – 4 ⇒ m = – 6
∴ l = 2 and m = – 6
∴ The separate equation of the straight lines are
3x – 4y – 6 = 0 and 3x – 4y + 2 = 0
The distance between the parallel lines is given by
∴ The given pair of straight lines are parallel and the distance between them is \(\frac{8}{5}\) units
The given equation of the pair of straight line is
9x 2 – 24xy + by 2 – 12x + 16y – 12 = 0 ………… (1)
9x 2 – 24xy + 16y 2 = 9x 2 – 12xy – 12xy + 16y 2
= 3x (3x – 4y) – 4y (3x – 4y)
= (3x – 4y) (3x – 4y)
Let the separate equation of the straight lines be
3x – 4y + 1 = 0 and 3x – 4y + m = 0
9x 2 – 24xy + 16y 2 – 12x + 16y – 12
= (3x – 4y + l) ( 3x – 4y + m )
Comparing the coefficients of x, y and constant terms on both sides
3l + 3m = – 12
l + m = – 4 ……….. (2)
– 4l – 4m = 16
l + m = – 4 ………… (3)
lm = – 12 ……….. (4)
(l – m) 2 = (l + m) 2 – 4lm
= (- 4) 2 – 4 × – 12
= 16 + 48 = 64
l – m = √64 = 8
l – m = 8 ………… (5)
Solving equations (2) and (5 ), we have
(2) ⇒ 2 + m = – 4 ⇒ m = – 6
∴ l = 2 and m = – 6
∴ The separate equation of the straight lines are
3x – 4y – 6 = 0 and 3x – 4y + 2 = 0
The distance between the parallel lines is given by
∴ The given pair of straight lines are parallel and the distance between them is \(\frac{8}{5}\) units
The given equation of pair of straight lines is
4x 2 + 4xy + y 2 – 6x – 3y – 4 = 0 ………. (1)
4x 2 + 4xy + y 2 = (2x + y) 2
Let the separate equation of the lines be
2x + y + l = 0 ……….. (2)
2x + y + m = 0 ………. (3)
4x 2 + 4xy + y 2 – 6x – 3y – 4 = (2x + y + l) (2x + y + m)
Comparing the coefficients of x, y and constant terms on both sides we have
2l + 2m = – 6
l + m = – 3 ……… (4)
l + m = – 3 ……… (5)
l m = – 4 ……… (6)
(l – m) 2 = (l + m) 2 – 4lm
(l – m ) 2 = (- 3) 2 – 4 × – 4
(l – m) 2 = 9 + 16 = 25
l – m = 5 ………… (7)
Solving equations (4) and (7)
(4) ⇒ l + m = – 3 ⇒ m = – 4
∴ The separate equation of the straight lines are
2x + y + 1 =0 and 2x + y – 4 = 0
The distance between the parallel lines is
d = \(\frac{5}{\sqrt{5}}\) = \(\sqrt{5}\)
∴ The given equation represents a pair of parallel straight lines and the distance between the parallel lines is \(\sqrt{5}\) units.
The given equation of pair of straight lines is
4x 2 + 4xy + y 2 – 6x – 3y – 4 = 0 ………. (1)
4x 2 + 4xy + y 2 = (2x + y) 2
Let the separate equation of the lines be
2x + y + l = 0 ……….. (2)
2x + y + m = 0 ………. (3)
4x 2 + 4xy + y 2 – 6x – 3y – 4 = (2x + y + l) (2x + y + m)
Comparing the coefficients of x, y and constant terms on both sides we have
2l + 2m = – 6
l + m = – 3 ……… (4)
l + m = – 3 ……… (5)
l m = – 4 ……… (6)
(l – m) 2 = (l + m) 2 – 4lm
(l – m ) 2 = (- 3) 2 – 4 × – 4
(l – m) 2 = 9 + 16 = 25
l – m = 5 ………… (7)
Solving equations (4) and (7)
(4) ⇒ l + m = – 3 ⇒ m = – 4
∴ The separate equation of the straight lines are
2x + y + 1 =0 and 2x + y – 4 = 0
The distance between the parallel lines is
d = \(\frac{5}{\sqrt{5}}\) = \(\sqrt{5}\)
∴ The given equation represents a pair of parallel straight lines and the distance between the parallel lines is \(\sqrt{5}\) units.
The equation of the given pair of straight lines is
ax 2 + 2hxy + by 2 = 0 ……… (1)
Let m 1 and m 2, be the slopes of the separate straight lines.
Given that one of the straight lines of (1) bisects the angle between the coordinate axes.
∴ The angle made by that line with x-axis 45°.
Slope of that line m 1 = tan 45°
m 1 = 1
Squaring on both sides
(a + b) 2 = (- 2h) 2
(a + b) 2 = 4h 2
The equation of the given pair of straight lines is
ax 2 + 2hxy + by 2 = 0 ……… (1)
Let m 1 and m 2, be the slopes of the separate straight lines.
Given that one of the straight lines of (1) bisects the angle between the coordinate axes.
∴ The angle made by that line with x-axis 45°.
Slope of that line m 1 = tan 45°
m 1 = 1
Squaring on both sides
(a + b) 2 = (- 2h) 2
(a + b) 2 = 4h 2
The equations of the given pair of straight lines are
x 2 – 2kxy – y 2 = 0 ………… (1)
x 2 – 2lxy – y 2 = 0 ………… (2)
Given that the pair x 2 – 2kxy – y 2 = 0 bisects the angle between the pair x 2 – 2lxy – y 2 = 0
∴ The equation of the bisector of the pair
x 2 – 2lxy – y 2 = 0 is the pair x 2 – 2kxy – y 2 = 0
The equation of the bisector of x 2 – 2lxy – y 2 = 0 is
Equation (3) and Equation (1) represents the same straight lines. ∴ The coefficients are proportional.
To show that the pair x 2 – 2lxy – y 2 = 0 bisects the angle between the pair x 2 – 2kxy – y 2 = 0, it is enough to prove the equation of the bisector of x 2 – 2kxy – y 2 = 0 is x 2 – 2lxy – y 2 = 0
The equation of the bisector of x 2 – 2kxy – y 2 = 0 is
x 2 – y 2 = 2lxy.
x 2 – 2lxy – y 2 = 0
∴ The pair x 2 – 2lxy – y 2 = 0 bisects the angle between the pair x 2 – 2kxy – y 2 = 0
The equations of the given pair of straight lines are
x 2 – 2kxy – y 2 = 0 ………… (1)
x 2 – 2lxy – y 2 = 0 ………… (2)
Given that the pair x 2 – 2kxy – y 2 = 0 bisects the angle between the pair x 2 – 2lxy – y 2 = 0
∴ The equation of the bisector of the pair
x 2 – 2lxy – y 2 = 0 is the pair x 2 – 2kxy – y 2 = 0
The equation of the bisector of x 2 – 2lxy – y 2 = 0 is
Equation (3) and Equation (1) represents the same straight lines. ∴ The coefficients are proportional.
To show that the pair x 2 – 2lxy – y 2 = 0 bisects the angle between the pair x 2 – 2kxy – y 2 = 0, it is enough to prove the equation of the bisector of x 2 – 2kxy – y 2 = 0 is x 2 – 2lxy – y 2 = 0
The equation of the bisector of x 2 – 2kxy – y 2 = 0 is
x 2 – y 2 = 2lxy.
x 2 – 2lxy – y 2 = 0
∴ The pair x 2 – 2lxy – y 2 = 0 bisects the angle between the pair x 2 – 2kxy – y 2 = 0
Homogenizing the given equations 3x 2 + 5xy – 3y 2 + 2x + 3y = 0 and 3x – 2y – 1 = 0
(i.e) 3x – 2y = 1.
We get (3x 2 + 5xy – 3y 2 ) + (2x + 3y)( 1) = 0
(i.e) (3x 2 + 5xy – 3y 2 ) + (2x + 3y)(3x – 2y) = 0
3x 2 + 5xy – 3y 2 + bx 2 – 4xy + 9xy – 6y 2 = 0
9x 2 + 10xy – 9y 2 = 0
Coefficient of x 2 + coefficient of y 2 = 9 – 9 = 0
⇒ The pair of straight lines are at right angles.
Homogenizing the given equations 3x 2 + 5xy – 3y 2 + 2x + 3y = 0 and 3x – 2y – 1 = 0
(i.e) 3x – 2y = 1.
We get (3x 2 + 5xy – 3y 2 ) + (2x + 3y)( 1) = 0
(i.e) (3x 2 + 5xy – 3y 2 ) + (2x + 3y)(3x – 2y) = 0
3x 2 + 5xy – 3y 2 + bx 2 – 4xy + 9xy – 6y 2 = 0
9x 2 + 10xy – 9y 2 = 0
Coefficient of x 2 + coefficient of y 2 = 9 – 9 = 0
⇒ The pair of straight lines are at right angles.
(4) 3x 2 – y 2 = 0
Explaination:
Let the point be (h,k)
Equation of y axis is x = 0
Given the distance of the point from y – axis is
\(\frac{1}{2}\) × distance of the point from the origin Distance of the point (h, k) from the y- axis is
= \(\frac{\mathrm{h}}{\sqrt{1^{2}}}\) = h
The distance of the point P (h, k) from the origin
Squaring on both sides
4h 2 = h 2 + k 2
4h 2 – h 2 – k 2 = 0
3h 2 – k 2 = o
The locus of ( h, k ) is 3x 2 – y 2 = 0
(4) 3x 2 – y 2 = 0
Explaination:
Let the point be (h,k)
Equation of y axis is x = 0
Given the distance of the point from y – axis is
\(\frac{1}{2}\) × distance of the point from the origin Distance of the point (h, k) from the y- axis is
= \(\frac{\mathrm{h}}{\sqrt{1^{2}}}\) = h
The distance of the point P (h, k) from the origin
Squaring on both sides
4h 2 = h 2 + k 2
4h 2 – h 2 – k 2 = 0
3h 2 – k 2 = o
The locus of ( h, k ) is 3x 2 – y 2 = 0
(4) y 2 = 4ax
Explaination:
y 2 = 4ax ⇒ Equation that satisfies the given point (at 2, 2at)
(4) y 2 = 4ax
Explaination:
y 2 = 4ax ⇒ Equation that satisfies the given point (at 2, 2at)
(3) (1, 2)
Explaination:
The point that satisfies the given equations (0, 0) ⇒ 17 ≠ 0
(-2, 3) ⇒ 3 (4) + 3 (9) + 16 – 36 + 17 ≠ 0
(1, 2) ⇒ 3 + 3 (4) – 8 (1) – 12 (2) + 17
32 – 32 = 0, 0 = 0
(3) (1, 2)
Explaination:
The point that satisfies the given equations (0, 0) ⇒ 17 ≠ 0
(-2, 3) ⇒ 3 (4) + 3 (9) + 16 – 36 + 17 ≠ 0
(1, 2) ⇒ 3 + 3 (4) – 8 (1) – 12 (2) + 17
32 – 32 = 0, 0 = 0
(4) 3
Explaination:
Given the point (8, – 5) lies on the locus
(4) 3
Explaination:
Given the point (8, – 5) lies on the locus
(3) α + 3β = 11
Explaination:
The equation of the straight line joining the points (2, 3) and (-1, 4) is
x – 2 = – 3(y – 3)
x – 2 = – 3y + 9
x + 3y – 2 – 9 = 0
x + 3y – 11 = 0
Given (α, P) lies on this line
α + 3β – 11 = 0
(3) α + 3β = 11
Explaination:
The equation of the straight line joining the points (2, 3) and (-1, 4) is
x – 2 = – 3(y – 3)
x – 2 = – 3y + 9
x + 3y – 2 – 9 = 0
x + 3y – 11 = 0
Given (α, P) lies on this line
α + 3β – 11 = 0
(2) \(\frac{1}{2}\), – 2
Explaination:
The equation of the given line is
3x – y + 5 = 0 ………… (1)
the slope of the line (1) is m = tan θ
Angle made by the required line = θ + 45°
where tan θ = 3
Required slope m 1 = tan (θ + 45° )
(2) \(\frac{1}{2}\), – 2
Explaination:
The equation of the given line is
3x – y + 5 = 0 ………… (1)
the slope of the line (1) is m = tan θ
Angle made by the required line = θ + 45°
where tan θ = 3
Required slope m 1 = tan (θ + 45° )
(2) x + y – 2 = 0
Explaination:
Let OAB be the isosceles triangle formed in the first quadrant. Let OA = OB = a
In the right angle A OAB
AB 2 = OA 2 + OB 2
AB 2 = a 2 + a 2 = 2a 2
AB = √2a
(2) x + y – 2 = 0
Explaination:
Let OAB be the isosceles triangle formed in the first quadrant. Let OA = OB = a
In the right angle A OAB
AB 2 = OA 2 + OB 2
AB 2 = a 2 + a 2 = 2a 2
AB = √2a
(4) x – y + 3 = 0
Explaination:
ABCD is a quadrilateral in which the sides AD and BC are parallel. Draw AE and DF perpendicular to BC. ADFE is a square with sides
AD = AE = EF = DF = 2
Area of ADFE = 2 × 2 = 4
In ∆ AEB, BE = 1, AE = 2
∴ Area of ∆ AEB = \(\frac{1}{2}\) × 1 × 2 = 1
Similarly Area of ∆ DFC = 1
∆ Area of the quadrilateral A B C D = 4 + 1 + 1 = 6
Given the line through the vertex (- 1, 2 ) divides the quadrilateral ABCD into two half.
Let E (x, 4) is the point on the line BC such that the line DE divides the area of the quadrilateral ABCD into two half.
∴ Area of ∆ EDC = 3
\(\frac{1}{2}\) × EC × height = 3
\(\frac{1}{2}\) × (x + 2) × 2 = 3
x + 2 = 3 ⇒ x = 1
∴ The coordinates of E are (1, 4)
The equation of the line joining the points (- 1, 2) and (1, 4) is
x – 1 = y – 4
x – y + 3 = 0
(4) x – y + 3 = 0
Explaination:
ABCD is a quadrilateral in which the sides AD and BC are parallel. Draw AE and DF perpendicular to BC. ADFE is a square with sides
AD = AE = EF = DF = 2
Area of ADFE = 2 × 2 = 4
In ∆ AEB, BE = 1, AE = 2
∴ Area of ∆ AEB = \(\frac{1}{2}\) × 1 × 2 = 1
Similarly Area of ∆ DFC = 1
∆ Area of the quadrilateral A B C D = 4 + 1 + 1 = 6
Given the line through the vertex (- 1, 2 ) divides the quadrilateral ABCD into two half.
Let E (x, 4) is the point on the line BC such that the line DE divides the area of the quadrilateral ABCD into two half.
∴ Area of ∆ EDC = 3
\(\frac{1}{2}\) × EC × height = 3
\(\frac{1}{2}\) × (x + 2) × 2 = 3
x + 2 = 3 ⇒ x = 1
∴ The coordinates of E are (1, 4)
The equation of the line joining the points (- 1, 2) and (1, 4) is
x – 1 = y – 4
x – y + 3 = 0
(2) 5, 5
Explaination:
Let the given points be A (1, 2) and B (3, 4) P( h,k )be a point in the plane such that PA = PB
Squaring on both the sides
(h – 1) 2 + (k – 2) 2 = (3 – h) 2 + (k – 4) 2
h 2 – 2h + 1 + k 2 – 4k + 4 = 9 – 6h + h 2 + k 2 – 8k + 16
– 2h – 4k + 5 = – 6h – 8k + 25
6h + 8k – 2h – 4k + 5 – 25 = 0
4h + 4k – 20 = 0
h + k – 5 = 0
The locus of (h, k)is x + y – 5 = 0 which is the perpendicular bisector of A and B
x + y = 5
\(\frac{x}{5}+\frac{y}{5}\) = 1
x – intercept = 5, y – intercept = 5
(2) 5, 5
Explaination:
Let the given points be A (1, 2) and B (3, 4) P( h,k )be a point in the plane such that PA = PB
Squaring on both the sides
(h – 1) 2 + (k – 2) 2 = (3 – h) 2 + (k – 4) 2
h 2 – 2h + 1 + k 2 – 4k + 4 = 9 – 6h + h 2 + k 2 – 8k + 16
– 2h – 4k + 5 = – 6h – 8k + 25
6h + 8k – 2h – 4k + 5 – 25 = 0
4h + 4k – 20 = 0
h + k – 5 = 0
The locus of (h, k)is x + y – 5 = 0 which is the perpendicular bisector of A and B
x + y = 5
\(\frac{x}{5}+\frac{y}{5}\) = 1
x – intercept = 5, y – intercept = 5
Let the equation of the required line be
y = mx + c ……….. (1)
Given Slope m = 2
(1) ⇒ y = 2x + c
2x – y + c = 0 ……….. (2)
The length of the perpendicular from (0, 0) to line (2)
Substituting for c in equation (2), we have
2x – y + 5 = 0
Let the equation of the required line be
y = mx + c ……….. (1)
Given Slope m = 2
(1) ⇒ y = 2x + c
2x – y + c = 0 ……….. (2)
The length of the perpendicular from (0, 0) to line (2)
Substituting for c in equation (2), we have
2x – y + 5 = 0
(1) x + 5y ± 5√2 = 0
Explaination:
The equation of the given line is 5x – y = 0 —— (1)
The equation of any perpendicular to line (1) is
-x – 5y + k = 0
x + 5y – k = 0 —– (2)
x + 5y = k
x- intercept = k, y – intercept = \(\frac{\mathrm{k}}{5}\)
The line (2) intersect the x-axis at A and y-axis at B
∴ A is (k, 0) and B is \(\left(0, \frac{k}{5}\right)\)
OA = k and OB = \(\frac{\mathrm{k}}{5}\)
Area of ∆ OBA = \(\frac{1}{2}\) × OA × OB
∴ The required equation is x + 5y = ± 5√2
(1) x + 5y ± 5√2 = 0
Explaination:
The equation of the given line is 5x – y = 0 —— (1)
The equation of any perpendicular to line (1) is
-x – 5y + k = 0
x + 5y – k = 0 —– (2)
x + 5y = k
x- intercept = k, y – intercept = \(\frac{\mathrm{k}}{5}\)
The line (2) intersect the x-axis at A and y-axis at B
∴ A is (k, 0) and B is \(\left(0, \frac{k}{5}\right)\)
OA = k and OB = \(\frac{\mathrm{k}}{5}\)
Area of ∆ OBA = \(\frac{1}{2}\) × OA × OB
∴ The required equation is x + 5y = ± 5√2
(2) x + y – 5 = 0
Explaination:
The equation of the given line is
x – y + 5 = 0 ………. (1)
To find the point of intersection with y – axis, Put x = 0 in (1)
0 – y + 5 = 0 ⇒ y = 5
∴ The point of intersection is ( 0, 5)
The equation any line perpendicular to line (1) is
– x – y + k = 0
x + y – k = 0
This line passes through the point (0,5)
0 + 5 – k = 0 ⇒ k = 5
∴ The equation of the required line is
x + y – 5 = 0
(2) x + y – 5 = 0
Explaination:
The equation of the given line is
x – y + 5 = 0 ………. (1)
To find the point of intersection with y – axis, Put x = 0 in (1)
0 – y + 5 = 0 ⇒ y = 5
∴ The point of intersection is ( 0, 5)
The equation any line perpendicular to line (1) is
– x – y + k = 0
x + y – k = 0
This line passes through the point (0,5)
0 + 5 – k = 0 ⇒ k = 5
∴ The equation of the required line is
x + y – 5 = 0
(3) √6
Explaination:
∆ ABC is an equilateral triangle. Vertex A is (2, 3)
Equation of the base BC is x + y = 2
Draw AD ⊥ BC. By the property of the equilateral triangle, D is the midpoint of BC
BD = DC
Let AB = a
AD = length of the perpendicular from A (2, 3) to the line x + y – 2 = 0
(3) √6
Explaination:
∆ ABC is an equilateral triangle. Vertex A is (2, 3)
Equation of the base BC is x + y = 2
Draw AD ⊥ BC. By the property of the equilateral triangle, D is the midpoint of BC
BD = DC
Let AB = a
AD = length of the perpendicular from A (2, 3) to the line x + y – 2 = 0
(4) \(\left(\frac{2}{5}, \frac{3}{5}\right)\)
Explaination:
The equation of the given line is
(p + 2q)x + (p – 3q)y = p – q ……….. (1)
(4) \(\left(\frac{2}{5}, \frac{3}{5}\right)\)
Explaination:
The equation of the given line is
(p + 2q)x + (p – 3q)y = p – q ……….. (1)
(2) (4, 1)
Explaination:
Let P (h, k) be a point on the line
2x – 3y – 5 = 0
Then 2h – 3k – 5 = 0 ……….. (1)
Let A (1, 2) and B ( 3, 4) be the given points such that PA = PB
Squaring on both sides
(h – 1) 2 + (k – 2) 2 = (h – 3) 2 + (k – 4) 2
h 2 – 2h + 1 + k 2 – 4k + 4 = h 2 – 6h + 9 + k 2 – 8k + 16
-2h + 1 – 4k + 4 = – 6h + 9 – 8k + 16
-2h – 4k + 5 = – 6h – 8k + 25
6h – 2h + 8k – 4k + 5 – 25 = 0
4h + 4k – 20 = 0
h + k – 5 = 0 …………. (2)
To find P (h, k), Solve (1) and (2)
(2) ⇒ h + 1 – 5 = 0
h – 4 = 0 ⇒ h = 4
∴ The required point p(h, k) is (4, 1)
(2) (4, 1)
Explaination:
Let P (h, k) be a point on the line
2x – 3y – 5 = 0
Then 2h – 3k – 5 = 0 ……….. (1)
Let A (1, 2) and B ( 3, 4) be the given points such that PA = PB
Squaring on both sides
(h – 1) 2 + (k – 2) 2 = (h – 3) 2 + (k – 4) 2
h 2 – 2h + 1 + k 2 – 4k + 4 = h 2 – 6h + 9 + k 2 – 8k + 16
-2h + 1 – 4k + 4 = – 6h + 9 – 8k + 16
-2h – 4k + 5 = – 6h – 8k + 25
6h – 2h + 8k – 4k + 5 – 25 = 0
4h + 4k – 20 = 0
h + k – 5 = 0 …………. (2)
To find P (h, k), Solve (1) and (2)
(2) ⇒ h + 1 – 5 = 0
h – 4 = 0 ⇒ h = 4
∴ The required point p(h, k) is (4, 1)
(1) (- 3, – 2)
Explaination:
The equation of the given line is y = -x
x + y = 0 ………… (1)
The coordinates of the image of the point (x 1, y 1 ) with respect to the line ax + by + c = 0 are given by
∴ The coordinates of the image of the point (2, 3) with respect to the line x + y = 0 are given by
x – 2 = y – 3 = – 5
x – 2 = -5 ⇒ x = -5 + 2 = -3
y – 3 = -5 ⇒ y = -5 + 3 = -2
∴ The required point is (-3, -2)
(1) (- 3, – 2)
Explaination:
The equation of the given line is y = -x
x + y = 0 ………… (1)
The coordinates of the image of the point (x 1, y 1 ) with respect to the line ax + by + c = 0 are given by
∴ The coordinates of the image of the point (2, 3) with respect to the line x + y = 0 are given by
x – 2 = y – 3 = – 5
x – 2 = -5 ⇒ x = -5 + 2 = -3
y – 3 = -5 ⇒ y = -5 + 3 = -2
∴ The required point is (-3, -2)
(3) \(\frac{12}{5}\)
Explaination:
The equation of the given line is \(\frac{x}{3}-\frac{y}{4}\) = 1
\(\frac{4 x-3 y}{12}\) = 14x – 3y = 12
4x – 3y – 12 = 0
The length of the perpendicular from (0, 0) to the line 4x – 3y – 12 = 0 is
(3) \(\frac{12}{5}\)
Explaination:
The equation of the given line is \(\frac{x}{3}-\frac{y}{4}\) = 1
\(\frac{4 x-3 y}{12}\) = 14x – 3y = 12
4x – 3y – 12 = 0
The length of the perpendicular from (0, 0) to the line 4x – 3y – 12 = 0 is
(2) \(\frac{9}{2}\)
Explaination:
The equation of the given line is
2x – 3y + 1 = 0 …………. (1)
The equation of any line perpendicular to (1) is
-3x – 2y + k = 0
3x + 2y – k = 0 ………… (2)
This line passes through the point (1, 3)
∴ (2) ⇒ 3 × 1 + 2 × 3 – k = 0
3 + 6 – k = 0 ⇒ k = 9
∴ (2) ⇒ 3x + 2y – 9 = 0 ………. (3)
2y = – 3x + 9
y = \(-\frac{3}{2} x+\frac{9}{2}\)
∴ The required y – intercept is \(\frac{9}{2}\)
(2) \(\frac{9}{2}\)
Explaination:
The equation of the given line is
2x – 3y + 1 = 0 …………. (1)
The equation of any line perpendicular to (1) is
-3x – 2y + k = 0
3x + 2y – k = 0 ………… (2)
This line passes through the point (1, 3)
∴ (2) ⇒ 3 × 1 + 2 × 3 – k = 0
3 + 6 – k = 0 ⇒ k = 9
∴ (2) ⇒ 3x + 2y – 9 = 0 ………. (3)
2y = – 3x + 9
y = \(-\frac{3}{2} x+\frac{9}{2}\)
∴ The required y – intercept is \(\frac{9}{2}\)
(1) k = 3
Explaination:
The equation of the given lines are
x + (2k – 7)y + 3 = 0 ……….. (1)
3kx + 9y – 5 = 0 …………. (2)
Slope of line (1), m 1 = \(-\frac{1}{2 k-7}\)
Slope of line (2), m 2 = \(-\frac{3 \mathbf{k}}{9}\)
Given that lines (1) and (2) are perpendicular
∴ m 1 m 2 = -1
k = -3(2k – 7)
k = -6k + 21
6k + k = 21
⇒ 7k = 21
k = \(\frac{21}{7}\) = 3
(1) k = 3
Explaination:
The equation of the given lines are
x + (2k – 7)y + 3 = 0 ……….. (1)
3kx + 9y – 5 = 0 …………. (2)
Slope of line (1), m 1 = \(-\frac{1}{2 k-7}\)
Slope of line (2), m 2 = \(-\frac{3 \mathbf{k}}{9}\)
Given that lines (1) and (2) are perpendicular
∴ m 1 m 2 = -1
k = -3(2k – 7)
k = -6k + 21
6k + k = 21
⇒ 7k = 21
k = \(\frac{21}{7}\) = 3
(2) 16 sq. units
Explanation:
OABC is a square in which the vertex O lies on the origin and the side AB along the line 4x + 3y – 20 = 0
OA = length of the perpendicular from (0, 0) to the line 4x + 3y – 20 = 0
∴ The length of the side of the square is 4 units.
∴ Area of the square = 4 × 4 = 16 sq. units
(2) 16 sq. units
Explanation:
OABC is a square in which the vertex O lies on the origin and the side AB along the line 4x + 3y – 20 = 0
OA = length of the perpendicular from (0, 0) to the line 4x + 3y – 20 = 0
∴ The length of the side of the square is 4 units.
∴ Area of the square = 4 × 4 = 16 sq. units
(1) –\(\frac{6}{7}\)
Explaination:
The equation of the given pair of lines is
6x 2 + 41xy – 7y 2 = 0 ……….. (1)
Let m 1, m 2 be slopes of the individual lines.
Given that the individual lines make angles α and β with x – axis.
∴ m 1 = tan α and m 2 = tan β
we have m 1 m 2 = \(\frac{a}{b}\) = \(\frac{6}{-7}\)
∴ tan α tan β = \(-\frac{6}{7}\)
(1) –\(\frac{6}{7}\)
Explaination:
The equation of the given pair of lines is
6x 2 + 41xy – 7y 2 = 0 ……….. (1)
Let m 1, m 2 be slopes of the individual lines.
Given that the individual lines make angles α and β with x – axis.
∴ m 1 = tan α and m 2 = tan β
we have m 1 m 2 = \(\frac{a}{b}\) = \(\frac{6}{-7}\)
∴ tan α tan β = \(-\frac{6}{7}\)
(3) \(\frac{1}{2}\)
Explaination:
The equation of the given pair of straight lines is
x 2 – 4y 2 = 0
x 2 – 4y 2 = (x + 2y) (x – 2y)
∴ The separate equations of the straight lines are
x + 2y = 0 ………… (1)
x – 2y = 0 …………. (2)
The line x = a intersect the line (1) at A.
a + 2y = 0 ⇒ y = \(-\frac{a}{2}\)
∴ A is \(\left(a,-\frac{a}{2}\right)\)
The line x = a intersect the line (2) at B.
(3) \(\frac{1}{2}\)
Explaination:
The equation of the given pair of straight lines is
x 2 – 4y 2 = 0
x 2 – 4y 2 = (x + 2y) (x – 2y)
∴ The separate equations of the straight lines are
x + 2y = 0 ………… (1)
x – 2y = 0 …………. (2)
The line x = a intersect the line (1) at A.
a + 2y = 0 ⇒ y = \(-\frac{a}{2}\)
∴ A is \(\left(a,-\frac{a}{2}\right)\)
The line x = a intersect the line (2) at B.
(1) -3
Explaination:
The given pair of straight lines is 6x 2 – xy – 4cy 2 = 0 ………. (1)
One of the separate equation is 3x + 4y = 0 ……… (2)
Let the separate equation of the other line be ax + by = 0 ………. (3)
∴ (ax + by) (3x + 4y) = 6x 2 – xy + 4cy 2
3ax 2 + 4axy + 3bxy + 4by 2 = 6x 2 – xy + 4cy 2
Equating the coefficient of y 2 on both sides
4b = 4c ⇒ c = b
Equating the coefficient of x 2 on both sides
3a = 6
⇒ a = \(\frac{6}{3}\) = 2
Equating the coefficient of xy both sides
4a + 3b = -1
4 × 2 + 3b = -1 ⇒ 3b = -1 – 8
b = \(-\frac{9}{3}\) = -3
∴ c = -3
(1) -3
Explaination:
The given pair of straight lines is 6x 2 – xy – 4cy 2 = 0 ………. (1)
One of the separate equation is 3x + 4y = 0 ……… (2)
Let the separate equation of the other line be ax + by = 0 ………. (3)
∴ (ax + by) (3x + 4y) = 6x 2 – xy + 4cy 2
3ax 2 + 4axy + 3bxy + 4by 2 = 6x 2 – xy + 4cy 2
Equating the coefficient of y 2 on both sides
4b = 4c ⇒ c = b
Equating the coefficient of x 2 on both sides
3a = 6
⇒ a = \(\frac{6}{3}\) = 2
Equating the coefficient of xy both sides
4a + 3b = -1
4 × 2 + 3b = -1 ⇒ 3b = -1 – 8
b = \(-\frac{9}{3}\) = -3
∴ c = -3
(3) \(\frac{5}{9}\)
Explaination:
The equation of the given pair of lines is
x 2 -xy – 6y 2 = 0
a = 1, 2h = -1, b = -6
Given θ is the angle between the lines
(3) \(\frac{5}{9}\)
Explaination:
The equation of the given pair of lines is
x 2 -xy – 6y 2 = 0
a = 1, 2h = -1, b = -6
Given θ is the angle between the lines
(4) x sin θ + y(cos θ + 2) = 0
Explaination:
The equation of given pair of straight line is
x 2 + 2xy cot θ – y 2 = 0
x 2 + (2y cot θ)x + (-y 2 ) = 0
This is a quadratic equation in x. Solving for x, we have
(4) x sin θ + y(cos θ + 2) = 0
Explaination:
The equation of given pair of straight line is
x 2 + 2xy cot θ – y 2 = 0
x 2 + (2y cot θ)x + (-y 2 ) = 0
This is a quadratic equation in x. Solving for x, we have