Samacheer Class 11 Maths - Two Dimensional Analytical Geometry

(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

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

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

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

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 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

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

(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 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

(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 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

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.

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

(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 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) 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

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
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 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

(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

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 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)

(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 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

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.

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

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

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 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
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
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 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

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

(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) 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) 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

(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

(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

(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

(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)

(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 \(\frac{9}{2}\)

(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

(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.

(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

(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