How do you find the area of a triangle from its vertices?

For vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃), Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

How do you check whether three points are collinear?

Three points are collinear if the area of the triangle formed by them is 0.

What is the slope of a line and how is it found?

The slope measures the steepness of a line. For two points, slope m = (y₂ − y₁)/(x₂ − x₁), and m = tan θ, where θ is the angle the line makes with the positive x-axis.

What are the conditions for two lines to be parallel or perpendicular?

Two lines are parallel when their slopes are equal (m₁ = m₂), and perpendicular when the product of their slopes is −1 (m₁ × m₂ = −1).

What is the slope-intercept form of a straight line?

The slope-intercept form is y = mx + c, where m is the slope of the line and c is its y-intercept.

Coordinate Geometry Class 10 Samacheer Solutions — Book Back Answers

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Ex 5.1Area of a Triangle and Quadrilateral11 questions

Q.1 Find the area of the triangle formed by the points (i) (1, −1), (−4, 6) and (−3, −5) (ii) (−10, −4), (−8, −1) and (−3, −5). ▾

Area = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

(i) ½|1(6 − (−5)) + (−4)((−5) − (−1)) + (−3)((−1) − 6)| = ½|11 + 16 + 21| = 24 sq.units.

(ii) ½|(−10)((−1) − (−5)) + (−8)((−5) − (−4)) + (−3)((−4) − (−1))| = ½|−40 + 8 + 9| = 11.5 sq.units.

Q.2 Determine whether the sets of points are collinear: (i) (−1/2, 3), (−5, 6) and (−8, 8) (ii) (a, b+c), (b, c+a) and (c, a+b). ▾

Points are collinear when the area of the triangle they form is 0.

(i) ½|(−1/2)(6 − 8) + (−5)(8 − 3) + (−8)(3 − 6)| = ½|1 − 25 + 24| = 0 ⇒ collinear.

(ii) ½|a((c+a) − (a+b)) + b((a+b) − (b+c)) + c((b+c) − (c+a))| = ½|a(c−b) + b(a−c) + c(b−a)| = 0 ⇒ collinear.

Q.3 The vertices of the given triangles are taken in order and their areas are provided. In each case find the value of p: (i) (0, 0), (p, 8), (6, 2) with area 20 sq.units (ii) (p, p), (5, 6), (5, −2) with area 32 sq.units. ▾

(i) ½|0(8 − 2) + p(2 − 0) + 6(0 − 8)| = 20 ⇒ |2p − 48| = 40 ⇒ p = 44 (the textbook value; p = 4 also satisfies the area).

(ii) ½|p(6 − (−2)) + 5((−2) − p) + 5(p − 6)| = 32 ⇒ |8p − 40| = 64 ⇒ p = 13 (p = −3 also satisfies the area).

Q.4 Find the value of a for which the given points are collinear: (i) (2, 3), (4, a) and (6, −3) (ii) (a, 2−2a), (−a+1, 2a) and (−4−a, 6−2a). ▾

(i) Collinear ⇒ area 0: 2(a − (−3)) + 4((−3) − 3) + 6(3 − a) = 0 ⇒ −4a = 0 ⇒ a = 0.

(ii) Setting the area to 0 and simplifying gives 8a² + 4a − 4 = 0 ⇒ 2a² + a − 1 = 0 ⇒ (2a − 1)(a + 1) = 0 ⇒ a = 1/2 or a = −1.

Q.5 Find the area of the quadrilateral whose vertices are at (i) (−9, −2), (−8, −4), (2, 2) and (1, −3) (ii) (−9, 0), (−8, 6), (−1, −2) and (−6, −3). ▾

Using the shoelace formula with the vertices taken in order around the boundary.

(i) Area = ½|(x₁y₂ − x₂y₁) + (x₂y₃ − x₃y₂) + (x₃y₄ − x₄y₃) + (x₄y₁ − x₁y₄)| = 35 sq.units.

(ii) Applying the same formula = 34 sq.units.

Q.6 Find the value of k, if the area of the quadrilateral is 28 sq.units, whose vertices are taken in the order (−4, −2), (−3, k), (3, −2) and (2, 3). ▾

By the shoelace formula the area = ½|−7k + 21| = (7/2)|k − 3|.

Setting (7/2)|k − 3| = 28 ⇒ |k − 3| = 8 ⇒ k = 11 or k = −5.

Q.7 If the points A(−3, 9), B(a, b) and C(4, −5) are collinear and a + b = 1, then find a and b. ▾

Slope AC = (−5 − 9)/(4 − (−3)) = −2. Collinear ⇒ slope AB = −2: (b − 9)/(a + 3) = −2 ⇒ b = −2a + 3.

With a + b = 1: a + (−2a + 3) = 1 ⇒ −a = −2 ⇒ a = 2, then b = −1.

Therefore a = 2, b = −1.

Q.8 Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the mid-points of the sides AB, BC and AC respectively of triangle ABC. Find the coordinates of A, B and C, the area of triangle ABC, and compare it with the area of triangle PQR. ▾

From the mid-point relations: x_A + x_B = 22, x_B + x_C = 27, x_A + x_C = 19 (and similarly for y). Solving gives A(7, 7), B(15, 7), C(12, 1).

Area of triangle ABC = 24 sq.units; area of triangle PQR = 6 sq.units.

Therefore area(ABC) = 24 sq.units = 4 × area(PQR).

Q.9 A quadrilateral swimming pool with vertices E(−3, −5), F(6, −2), G(3, 7) and H(−6, 4) is surrounded by a concrete patio bounded by A(−4, −8), B(8, −4), C(6, 10) and D(−10, 6). Find the area of the patio. ▾

Patio area = area of the outer quadrilateral ABCD − area of the pool EFGH.

By the shoelace formula, area(ABCD) = 212 sq.units and area(EFGH) = 90 sq.units.

Therefore the patio area = 212 − 90 = 122 sq.units.

Q.10 A triangular shaped glass with vertices at A(−5, −4), B(1, 6) and C(7, −4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass with one coat? ▾

Area = ½|(−5)(6 − (−4)) + 1((−4) − (−4)) + 7((−4) − 6)| = ½|−50 + 0 − 70| = 60 sq.feet.

Buckets needed = 60 ÷ 6 = 10 buckets.

Q.11 Given the plotted points A(−5, 3), F(−2, 3), D(1, 3), G(−4.5, 0.5), E(1.5, 1), B(−4, −2) and C(2, −1), find the area of (i) triangle AGF (ii) triangle FED (iii) quadrilateral BCEG. ▾

(i) AGF with A(−5, 3), G(−4.5, 0.5), F(−2, 3): area = ½|(−5)(0.5 − 3) + (−4.5)(3 − 3) + (−2)(3 − 0.5)| = ½|12.5 − 5| = 3.75 sq.units.

(ii) FED with F(−2, 3), E(1.5, 1), D(1, 3): area = ½|(−2)(1 − 3) + 1.5(3 − 3) + 1(3 − 1)| = ½|4 + 2| = 3 sq.units.

(iii) Quadrilateral BCEG with B(−4, −2), C(2, −1), E(1.5, 1), G(−4.5, 0.5): by the shoelace formula = ½(27.75) = 13.88 sq.units.

Ex 5.2Inclination of a Line13 questions

Q.1 What is the slope of a line whose inclination with the positive direction of the x-axis is (i) 90° (ii) 0°? ▾

(i) Slope = tan 90°, which is undefined (a vertical line).

(ii) Slope = tan 0° = 0 (a horizontal line).

Q.2 What is the inclination of a line whose slope is (i) 0 (ii) 1? ▾

(i) Inclination θ = tan⁻¹(0) = 0°.

(ii) Inclination θ = tan⁻¹(1) = 45°.

Q.3 Find the slope of a line joining the points (i) (5, 5) with the origin (ii) (sin θ, −cos θ) and (−sin θ, cos θ). ▾

(i) Slope = (5 − 0)/(5 − 0) = 1.

(ii) Slope = (cos θ − (−cos θ))/(−sin θ − sin θ) = (2 cos θ)/(−2 sin θ) = −cot θ.

Q.4 What is the slope of a line perpendicular to the line joining A(5, 1) and P, where P is the mid-point of the segment joining (4, 2) and (−6, 4)? ▾

P = ((4 + (−6))/2, (2 + 4)/2) = (−1, 3).

Slope of AP = (3 − 1)/(−1 − 5) = 2/(−6) = −1/3, so the perpendicular slope is 3.

Q.5 Show that the given points are collinear: (−3, −4), (7, 2) and (12, 5). ▾

Slope of (−3, −4) and (7, 2) = (2 − (−4))/(7 − (−3)) = 6/10 = 3/5.

Slope of (7, 2) and (12, 5) = (5 − 2)/(12 − 7) = 3/5.

Since the slopes are equal and a point is common, the three points are collinear.

Q.6 If the three points (3, −1), (a, 3) and (1, −3) are collinear, find the value of a. ▾

Slope of (3, −1) and (1, −3) = (−3 − (−1))/(1 − 3) = −2/−2 = 1.

For collinearity, slope of (3, −1) and (a, 3) must also be 1: (3 − (−1))/(a − 3) = 4/(a − 3) = 1 ⇒ a − 3 = 4.

Therefore a = 7.

Q.7 The line through the points (−2, a) and (9, 3) has slope −1/2. Find the value of a. ▾

Slope = (3 − a)/(9 − (−2)) = (3 − a)/11 = −1/2.

⇒ 2(3 − a) = −11 ⇒ 6 − 2a = −11 ⇒ −2a = −17, so a = 17/2.

Q.8 The line through the points (−2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x. ▾

Slope of the first line = (8 − 6)/(4 − (−2)) = 2/6 = 1/3, so the perpendicular line has slope −3.

Slope of the second line = (24 − 12)/(x − 8) = 12/(x − 8) = −3 ⇒ x − 8 = −4, so x = 4.

Q.9 Show that the given points form a right-angled triangle: (i) A(1, −4), B(2, −3) and C(4, −7) (ii) L(0, 5), M(9, 12) and N(3, 14). ▾

(i) Slope AB = (−3 − (−4))/(2 − 1) = 1; slope AC = (−7 − (−4))/(4 − 1) = −1. Since AB × AC = 1 × (−1) = −1, AB ⊥ AC — right-angled at A. Yes.

(ii) Slope MN = (14 − 12)/(3 − 9) = −1/3; slope LN = (14 − 5)/(3 − 0) = 3. Since MN × LN = (−1/3)(3) = −1, MN ⊥ LN — right-angled at N. Yes.

Q.10 Show that the given points form a parallelogram: A(2.5, 3.5), B(10, −4), C(2.5, −2.5) and D(−5, 5). ▾

Slope AB = (−4 − 3.5)/(10 − 2.5) = −1; slope CD = (5 − (−2.5))/(−5 − 2.5) = −1 ⇒ AB ∥ CD.

Slope BC = (−2.5 − (−4))/(2.5 − 10) = −1/5; slope DA = (3.5 − 5)/(2.5 − (−5)) = −1/5 ⇒ BC ∥ DA.

Both pairs of opposite sides are parallel, so ABCD is a parallelogram.

Q.11 If the points A(2, 2), B(−2, −3), C(1, −3) and D(x, y) form a parallelogram, then find the value of x and y. ▾

In a parallelogram the diagonals bisect each other, so mid-point of AC = mid-point of BD.

Mid-point AC = ((2 + 1)/2, (2 + (−3))/2) = (3/2, −1/2). Equating with mid-point BD = ((−2 + x)/2, (−3 + y)/2): −2 + x = 3 and −3 + y = −1.

Therefore x = 5, y = 2.

Q.12 Let A(3, −4), B(9, −4), C(5, −7) and D(7, −7). Show that ABCD is a trapezium. ▾

Slope AB = (−4 − (−4))/(9 − 3) = 0; slope CD = (−7 − (−7))/(7 − 5) = 0 ⇒ AB ∥ CD.

Slope BC = (−7 − (−4))/(5 − 9) = 3/4; slope AD = (−7 − (−4))/(7 − 3) = −3/4 ⇒ BC and AD are not parallel.

Exactly one pair of opposite sides is parallel, so ABCD is a trapezium.

Q.13 A quadrilateral has vertices at A(−4, −2), B(5, −1), C(6, 5) and D(−7, 6). Show that the mid-points of its sides form a parallelogram. ▾

Mid-points: P(AB) = (1/2, −3/2), Q(BC) = (11/2, 2), R(CD) = (−1/2, 11/2), S(DA) = (−11/2, 2).

Slope PQ = (2 − (−3/2))/(11/2 − 1/2) = 0.7; slope SR = (11/2 − 2)/(−1/2 − (−11/2)) = 0.7 ⇒ PQ ∥ SR.

Slope QR = (11/2 − 2)/(−1/2 − 11/2) = −7/12; slope PS = (2 − (−3/2))/(−11/2 − 1/2) = −7/12 ⇒ QR ∥ PS.

Both pairs of opposite sides are parallel, so PQRS is a parallelogram.

Ex 5.3Straight Line14 questions

Q.1 Find the equation of a straight line passing through the mid-point of the line segment joining the points (1, −5) and (4, 2), and parallel to (i) the X-axis (ii) the Y-axis. ▾

Mid-point = ((1 + 4)/2, (−5 + 2)/2) = (5/2, −3/2).

(i) Parallel to the X-axis is a horizontal line through the mid-point: y = −3/2, i.e. 2y + 3 = 0.

(ii) Parallel to the Y-axis is a vertical line through the mid-point: x = 5/2, i.e. 2x − 5 = 0.

Q.2 The equation of a straight line is 2(x − y) + 5 = 0. Find its slope, inclination and intercept on the Y-axis. ▾

2(x − y) + 5 = 0 ⇒ 2x − 2y + 5 = 0 ⇒ y = x + 5/2.

Slope m = 1; inclination θ = tan⁻¹(1) = 45°; Y-intercept = 5/2.

Q.3 Find the equation of a line whose inclination is 30° and making an intercept −3 on the Y-axis. ▾

Slope m = tan 30° = 1/√3, and y-intercept c = −3, so y = (1/√3)x − 3.

Multiplying by √3: √3 y = x − 3√3, i.e. x − √3 y − 3√3 = 0.

Q.4 Find the slope and y-intercept of √3 x + (1 − √3) y = 3. ▾

Write in slope-intercept form: y = [−√3/(1 − √3)] x + 3/(1 − √3).

Slope = −√3/(1 − √3) = (rationalising) (3 + √3)/2.

Y-intercept = 3/(1 − √3) = (rationalising) −(3 + 3√3)/2.

Q.5 Find the value of a, if the line through (−2, 3) and (8, 5) is perpendicular to y = ax + 2. ▾

Slope of the line through the points = (5 − 3)/(8 − (−2)) = 2/10 = 1/5.

The line y = ax + 2 has slope a. For perpendicular lines (1/5)·a = −1, so a = −5.

Q.6 The hill in the form of a right triangle has its foot at (19, 3). The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and the top. ▾

Slope = tan 45° = 1. Through the foot (19, 3): y − 3 = 1·(x − 19) ⇒ y = x − 16.

Therefore x − y − 16 = 0.

Q.7 Find the equation of a line through the given pair of points: (i) (2, 2/3) and (−1/2, −2) (ii) (2, 3) and (−7, −1). ▾

(i) Slope = (−2 − 2/3)/(−1/2 − 2) = (−8/3)/(−5/2) = 16/15. Through (2, 2/3): y − 2/3 = (16/15)(x − 2) ⇒ 15y − 10 = 16x − 32 ⇒ 16x − 15y − 22 = 0.

(ii) Slope = (−1 − 3)/(−7 − 2) = (−4)/(−9) = 4/9. Through (2, 3): y − 3 = (4/9)(x − 2) ⇒ 9y − 27 = 4x − 8 ⇒ 4x − 9y + 19 = 0.

Q.8 A cat is located at the point (−6, −4) in the xy-plane. A bottle of milk is kept at (5, 11). The cat wishes to consume the milk travelling through the shortest possible distance. Find the equation of the path it needs to take. ▾

The shortest path is the straight line joining the two points.

Slope = (11 − (−4))/(5 − (−6)) = 15/11. Through (−6, −4): y + 4 = (15/11)(x + 6) ⇒ 11y + 44 = 15x + 90.

Therefore 15x − 11y + 46 = 0.

Q.9 If the vertices of triangle ABC are A(6, 2), B(−5, −1) and C(1, 9), then through the vertex A find the equation of (i) the median (ii) the altitude. ▾

(i) Median: mid-point of BC = ((−5 + 1)/2, (−1 + 9)/2) = (−2, 4). Line through A(6, 2) and (−2, 4): slope = (4 − 2)/(−2 − 6) = −1/4. y − 2 = (−1/4)(x − 6) ⇒ 4y − 8 = −(x − 6) ⇒ x + 4y − 14 = 0.

(ii) Altitude through A is perpendicular to BC. Slope of BC = (9 − (−1))/(1 − (−5)) = 10/6 = 5/3, so the altitude slope is −3/5. y − 2 = (−3/5)(x − 6) ⇒ 5y − 10 = −3x + 18 ⇒ 3x + 5y − 28 = 0.

Q.10 Find the equation of a straight line which has slope −5/4 and passes through the point (−1, 2). ▾

y − 2 = (−5/4)(x + 1) ⇒ 4y − 8 = −5x − 5.

Therefore 5x + 4y − 3 = 0.

Q.11 You are downloading a song. The percent y (in decimal form) of megabytes remaining to get downloaded in x seconds is given by y = −0.1x + 1. (i) Find the total MB of the song. (ii) After how many seconds will 75% of the song get downloaded? (iii) After how many seconds will the song be downloaded completely? ▾

(i) At x = 0 (no time elapsed), y = 1 represents the whole song remaining, so the total size corresponds to 1 (MB unit of the model).

(ii) 75% downloaded means 25% (= 0.25) remaining: 0.25 = −0.1x + 1 ⇒ 0.1x = 0.75 ⇒ x = 7.5 seconds.

(iii) Completely downloaded means y = 0: 0 = −0.1x + 1 ⇒ x = 10 seconds.

Q.12 Find the equation of a line whose intercepts on the x and y axes are given below: (i) 4, −6 (ii) −5, 3/4. ▾

Using intercept form x/a + y/b = 1.

(i) x/4 + y/(−6) = 1 ⇒ (×12) 3x − 2y = 12 ⇒ 3x − 2y − 12 = 0.

(ii) x/(−5) + y/(3/4) = 1 ⇒ x/(−5) + 4y/3 = 1 ⇒ (×15) −3x + 20y = 15 ⇒ 3x − 20y + 15 = 0.

Q.13 Find the intercepts made by the following lines on the coordinate axes: (i) 3x − 2y − 6 = 0 (ii) 4x + 3y + 12 = 0. ▾

(i) Put y = 0: 3x = 6 ⇒ x-intercept = 2. Put x = 0: −2y = 6 ⇒ y-intercept = −3. So x-intercept 2, y-intercept −3.

(ii) Put y = 0: 4x = −12 ⇒ x-intercept = −3. Put x = 0: 3y = −12 ⇒ y-intercept = −4. So x-intercept −3, y-intercept −4.

Q.14 Find the equation of a straight line (i) passing through (1, −4) and whose intercepts on the axes are in the ratio 2:5 (ii) passing through (−8, 4) and making equal intercepts on the coordinate axes. ▾

(i) Let the intercepts be 2k and 5k: x/(2k) + y/(5k) = 1 ⇒ 5x + 2y = 10k. Through (1, −4): 5 − 8 = 10k ⇒ k = −3/10, so 5x + 2y = −3 ⇒ 5x + 2y + 3 = 0.

(ii) Equal intercepts a: x/a + y/a = 1 ⇒ x + y = a. Through (−8, 4): −8 + 4 = a = −4, so x + y = −4 ⇒ x + y + 4 = 0.

Ex 5.4General Form of a Straight Line12 questions

Q.1 Find the slope of the following straight lines: (i) 5y − 3 = 0 (ii) 7x − 3 = 0 ▾

(i) 5y − 3 = 0 ⇒ y = 3/5, a horizontal line, so its slope is 0.

(ii) 7x − 3 = 0 ⇒ x = 3/7, a vertical line, so its slope is undefined.

Q.2 Find the slope of the line which is (i) parallel to y = 0.7x − 11 (ii) perpendicular to the line x = −11. ▾

(i) Parallel lines have equal slopes, so the slope is 0.7.

(ii) x = −11 is a vertical line (undefined slope); a line perpendicular to it is horizontal, so its slope is 0.

Q.3 Check whether the given lines are parallel or perpendicular: (i) x/3 + y/4 + 1/7 = 0 and 2x/3 + y/2 + 1/10 = 0 (ii) 5x + 23y + 14 = 0 and 23x − 5y + 9 = 0 ▾

(i) Slope of x/3 + y/4 + 1/7 = 0 is −(1/3)/(1/4) = −4/3; slope of 2x/3 + y/2 + 1/10 = 0 is −(2/3)/(1/2) = −4/3. The slopes are equal, so the lines are parallel.

(ii) Slope of 5x + 23y + 14 = 0 is −5/23; slope of 23x − 5y + 9 = 0 is 23/5. Their product is (−5/23)(23/5) = −1, so the lines are perpendicular.

Q.4 If the straight lines 12y = −(p + 3)x + 12 and 12x − 7y = 16 are perpendicular, then find the value of p. ▾

From 12y = −(p + 3)x + 12 ⇒ y = −((p + 3)/12)x + 1, so slope m₁ = −(p + 3)/12.

From 12x − 7y = 16 ⇒ y = (12/7)x − 16/7, so slope m₂ = 12/7.

For perpendicular lines m₁·m₂ = −1: [−(p + 3)/12]·(12/7) = −1 ⇒ −(p + 3)/7 = −1 ⇒ p + 3 = 7.

Therefore p = 4.

Q.5 Find the equation of a straight line passing through the point P(−5, 2) and parallel to the line joining the points Q(3, −2) and R(−5, 4). ▾

Slope of QR = (4 − (−2))/(−5 − 3) = 6/(−8) = −3/4.

A line through P(−5, 2) parallel to QR has the same slope: y − 2 = (−3/4)(x + 5).

4(y − 2) = −3(x + 5) ⇒ 4y − 8 = −3x − 15.

Therefore 3x + 4y + 7 = 0.

Q.6 Find the equation of a line passing through (6, −2) and perpendicular to the line joining the points (6, 7) and (2, −3). ▾

Slope of the line joining (6, 7) and (2, −3) = (−3 − 7)/(2 − 6) = −10/−4 = 5/2.

The perpendicular slope is −2/5. Through (6, −2): y + 2 = (−2/5)(x − 6).

5(y + 2) = −2(x − 6) ⇒ 5y + 10 = −2x + 12.

Therefore 2x + 5y − 2 = 0.

Q.7 A(−3, 0), B(10, −2) and C(12, 3) are the vertices of triangle ABC. Find the equation of the altitude through A and the altitude through B. ▾

Altitude through A is perpendicular to BC. Slope of BC = (3 − (−2))/(12 − 10) = 5/2, so the altitude slope is −2/5. Through A(−3, 0): y = (−2/5)(x + 3) ⇒ 5y = −2x − 6 ⇒ 2x + 5y + 6 = 0.

Altitude through B is perpendicular to AC. Slope of AC = (3 − 0)/(12 − (−3)) = 1/5, so the altitude slope is −5. Through B(10, −2): y + 2 = −5(x − 10) ⇒ y + 2 = −5x + 50 ⇒ 5x + y − 48 = 0.

Q.8 Find the equation of the perpendicular bisector of the line joining the points A(−4, 2) and B(6, −4). ▾

Midpoint of AB = ((−4 + 6)/2, (2 + (−4))/2) = (1, −1).

Slope of AB = (−4 − 2)/(6 − (−4)) = −6/10 = −3/5, so the perpendicular bisector has slope 5/3.

Through (1, −1): y + 1 = (5/3)(x − 1) ⇒ 3(y + 1) = 5(x − 1) ⇒ 3y + 3 = 5x − 5.

Therefore 5x − 3y − 8 = 0.

Q.9 Find the equation of a straight line through the intersection of the lines 7x + 3y = 10 and 5x − 4y = 1 and parallel to the line 13x + 5y + 12 = 0. ▾

Solving 7x + 3y = 10 and 5x − 4y = 1: 4(7x + 3y) + 3(5x − 4y) = 40 + 3 ⇒ 43x = 43 ⇒ x = 1, then y = 1. Intersection (1, 1).

A line parallel to 13x + 5y + 12 = 0 is 13x + 5y + c = 0. Through (1, 1): 13 + 5 + c = 0 ⇒ c = −18.

Therefore 13x + 5y − 18 = 0.

Q.10 Find the equation of a straight line through the intersection of the lines 5x − 6y = 2 and 3x + 2y = 10 and perpendicular to the line 4x − 7y + 13 = 0. ▾

Solving 5x − 6y = 2 and 3x + 2y = 10: 3(3x + 2y) + (5x − 6y) = 30 + 2 ⇒ 14x = 32 ⇒ x = 16/7, then y = 11/7. Intersection (16/7, 11/7).

A line perpendicular to 4x − 7y + 13 = 0 (slope 4/7) has the form 7x + 4y + c = 0. Through (16/7, 11/7): 16 + 44/7 + c = 0 ⇒ c = −156/7.

Multiplying by 7: 49x + 28y − 156 = 0.

Q.11 Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x − 2y − 4 = 0 to the point of intersection of 7x − 3y = −12 and 2y = x + 3. ▾

First intersection: from 3x + y + 2 = 0, y = −3x − 2; substituting in x − 2y − 4 = 0 gives 7x = 0 ⇒ x = 0, y = −2. Point (0, −2).

Second intersection: from 2y = x + 3, x = 2y − 3; substituting in 7x − 3y = −12 gives 11y = 9 ⇒ y = 9/11, x = −15/11. Point (−15/11, 9/11).

Slope of the join = (9/11 + 2)/(−15/11 − 0) = (31/11)/(−15/11) = −31/15. Through (0, −2): y + 2 = (−31/15)x ⇒ 15y + 30 = −31x.

Therefore 31x + 15y + 30 = 0.

Q.12 Find the equation of a straight line through the point of intersection of the lines 8x + 3y = 18 and 4x + 5y = 9, and bisecting the line segment joining the points (5, −4) and (−7, 6). ▾

Solving 8x + 3y = 18 and 4x + 5y = 9: 2(4x + 5y) − (8x + 3y) = 18 − 18 ⇒ 7y = 0 ⇒ y = 0, then x = 9/4. Intersection (9/4, 0).

Midpoint of (5, −4) and (−7, 6) = (−1, 1). The required line passes through (9/4, 0) and (−1, 1).

Slope = (1 − 0)/(−1 − 9/4) = 1/(−13/4) = −4/13. Through (9/4, 0): y = (−4/13)(x − 9/4) ⇒ 13y = −4x + 9.

Therefore 4x + 13y − 9 = 0.

Q.1 Choose the correct answer: The area of the triangle formed by the points (−5, 0), (0, −5) and (5, 0) is (A) 0 sq.units (B) 25 sq.units (C) 5 sq.units (D) none of these. ▾

Area = ½ × base × height = ½ × 10 × 5 = 25. (B) 25 sq.units.

Q.2 Choose the correct answer: A man walks near a wall such that the distance between him and the wall is 10 units. Considering the wall as the Y-axis, the path travelled by the man is (A) x = 10 (B) y = 10 (C) x = 0 (D) y = 0. ▾

The distance from the Y-axis stays constant at 10, so the path is the vertical line (A) x = 10.

Q.3 Choose the correct answer: The straight line given by the equation x = 11 is (A) parallel to the X-axis (B) parallel to the Y-axis (C) passing through the origin (D) passing through the point (0, 11). ▾

x = 11 is a vertical line, so it is (B) parallel to the Y-axis.

Q.4 Choose the correct answer: If (5, 7), (3, p) and (6, 6) are collinear, then the value of p is (A) 3 (B) 6 (C) 9 (D) 12. ▾

Collinear ⇒ equal slopes: (p − 7)/(3 − 5) = (6 − 7)/(6 − 5) ⇒ (p − 7)/(−2) = −1 ⇒ p = 9. (C) 9.

Q.5 Choose the correct answer: The point of intersection of 3x − y = 4 and x + y = 8 is (A) (5, 3) (B) (2, 4) (C) (3, 5) (D) (4, 4). ▾

Adding the equations: 4x = 12 ⇒ x = 3, then y = 5. (C) (3, 5).

Q.6 Choose the correct answer: The slope of the line joining (12, 3) and (4, a) is 1/8. The value of a is (A) 1 (B) 4 (C) −5 (D) 2. ▾

Slope = (a − 3)/(4 − 12) = (a − 3)/(−8) = 1/8 ⇒ a − 3 = −1 ⇒ a = 2. (D) 2.

Q.7 Choose the correct answer: The slope of the line which is perpendicular to the line joining the points (0, 0) and (−8, 8) is (A) −1 (B) 1 (C) 1/3 (D) −8. ▾

Slope of the join = 8/(−8) = −1, so the perpendicular slope is 1. (B) 1.

Q.8 Choose the correct answer: If the slope of the line PQ is 1/3, then the slope of the perpendicular bisector of PQ is (A) 3 (B) −3 (C) 1/3 (D) 0. ▾

The perpendicular bisector is perpendicular to PQ, so its slope = −1 ÷ (1/3) = −3. (B) −3.

Q.9 Choose the correct answer: If A is a point on the Y-axis whose ordinate is 8 and B is a point on the X-axis whose abscissa is 5, then the equation of the line AB is (A) 8x + 5y = 40 (B) 8x − 5y = 40 (C) x = 8 (D) y = 5. ▾

A = (0, 8), B = (5, 0). Intercept form x/5 + y/8 = 1 ⇒ 8x + 5y = 40. (A) 8x + 5y = 40.

Q.10 Choose the correct answer: The equation of a line passing through the origin and perpendicular to the line 7x − 3y + 4 = 0 is (A) 7x − 3y + 4 = 0 (B) 3x − 7y + 4 = 0 (C) 3x + 7y = 0 (D) 7x − 3y = 0. ▾

Slope of the given line = 7/3, so the perpendicular slope is −3/7. Through the origin: y = (−3/7)x ⇒ 3x + 7y = 0. (C) 3x + 7y = 0.

Q.11 Choose the correct answer: Consider four straight lines (i) l₁: 3y = 4x + 5 (ii) l₂: 4y = 3x − 1 (iii) l₃: 4y + 3x = 7 (iv) l₄: 4x + 3y = 2. Which statement is true? (A) l₁ and l₂ are perpendicular (B) l₁ and l₄ are parallel (C) l₂ and l₄ are perpendicular (D) l₂ and l₃ are parallel. ▾

Slopes: l₁ = 4/3, l₂ = 3/4, l₃ = −3/4, l₄ = −4/3. Since l₂ × l₄ = (3/4)(−4/3) = −1, they are perpendicular. (C) l₂ and l₄ are perpendicular.

Q.12 Choose the correct answer: A straight line has the equation 8y = 4x + 21. Which of the following is true? (A) slope 0.5, y-intercept 2.6 (B) slope 5, y-intercept 1.6 (C) slope 0.5, y-intercept 1.6 (D) slope 5, y-intercept 2.6. ▾

y = (4/8)x + 21/8 = 0.5x + 2.625, so slope = 0.5 and y-intercept ≈ 2.6. (A) slope 0.5, y-intercept 2.6.

Q.13 Choose the correct answer: When proving that a quadrilateral is a trapezium, it is necessary to show (A) two sides are parallel (B) two parallel and two non-parallel sides (C) opposite sides are parallel (D) all sides are of equal length. ▾

A trapezium has exactly one pair of parallel sides — i.e. two parallel and two non-parallel sides. (B).

Q.14 Choose the correct answer: When proving that a quadrilateral is a parallelogram by using slopes, you must find (A) the slopes of two sides (B) the slopes of two pairs of opposite sides (C) the lengths of all sides (D) both the lengths and slopes of two sides. ▾

You must show both pairs of opposite sides are parallel — i.e. the slopes of two pairs of opposite sides. (B).

Q.15 Choose the correct answer: (2, 1) is the point of intersection of which two lines? (A) x − y − 3 = 0; 3x − y − 7 = 0 (B) x + y = 3; 3x + y = 7 (C) 3x + y = 3; x + y = 7 (D) x + 3y − 3 = 0; x − y − 7 = 0. ▾

Substituting (2, 1): x + y = 2 + 1 = 3 ✓ and 3x + y = 6 + 1 = 7 ✓. (B) x + y = 3; 3x + y = 7.

Ex 5.UEUnit Exercise10 questions

Q.1 PQRS is a rectangle formed by joining the points P(−1, −1), Q(−1, 4), R(5, 4) and S(5, −1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer. ▾

Mid-points: A(−1, 3/2), B(2, 4), C(5, 3/2), D(2, −1).

AB = BC = CD = DA = √(3² + 2.5²) = √15.25, so all four sides are equal, but the diagonals AC (length 6) and BD (length 5) are unequal — so it is not a square.

Therefore ABCD is a rhombus.

Q.2 The area of a triangle is 5 sq.units. Two of its vertices are (2, 1) and (3, −2). The third vertex is (x, y) where y = x + 3. Find the coordinates of the third vertex. ▾

Area = ½|2(−2 − y) + 3(y − 1) + x(1 − (−2))| = ½|3x + y − 7| = 5 ⇒ |3x + y − 7| = 10.

With y = x + 3: |4x − 4| = 10 ⇒ x = 7/2 or x = −3/2.

Third vertex = (7/2, 13/2) (or (−3/2, 3/2)).

Q.3 Find the area of the triangle formed by the lines 3x + y − 2 = 0, 5x + 2y − 3 = 0 and 2x − y − 3 = 0. ▾

Solving 3x + y = 2 and 2x − y = 3 gives (1, −1); substituting in 5x + 2y − 3 = 5 − 2 − 3 = 0 — so the third line also passes through (1, −1).

The three lines are concurrent, so the triangle has 0 sq.units area.

Q.4 If the vertices of a quadrilateral are at A(−5, 7), B(−4, k), C(−1, −6) and D(4, 5) and its area is 72 sq.units, find the value of k. ▾

By the shoelace formula the area = ½|−4k + 124| = 72 ⇒ |−4k + 124| = 144.

⇒ −4k + 124 = ±144 ⇒ k = −5 (or k = 67). Taking the textbook value, k = −5.

Q.5 Without using the distance formula, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram. ▾

Slope of (−2,−1)–(4,0) = 1/6; slope of (3,3)–(−3,2) = 1/6 ⇒ one pair of opposite sides is parallel.

Slope of (4,0)–(3,3) = −3; slope of (−3,2)–(−2,−1) = −3 ⇒ the other pair is parallel.

Both pairs of opposite sides are parallel, so the points form a parallelogram.

Q.6 Find the equations of the lines whose sum and product of intercepts are 1 and −6 respectively. ▾

Let the intercepts be a and b: a + b = 1, ab = −6 ⇒ a, b are roots of t² − t − 6 = 0 ⇒ (t − 3)(t + 2) = 0 ⇒ {3, −2}.

Intercept form x/a + y/b = 1 gives: x/3 + y/(−2) = 1 ⇒ 2x − 3y − 6 = 0, and x/(−2) + y/3 = 1 ⇒ 3x − 2y + 6 = 0.

Q.7 The owner of a milk store finds that he can sell 980 litres of milk each week at ₹14/litre and 1220 litres each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre? ▾

Treating demand q as linear in price p through (14, 980) and (16, 1220): slope = (1220 − 980)/(16 − 14) = 120.

q − 980 = 120(p − 14). At p = 17: q = 980 + 120(3) = 1340 litres.

Q.8 Find the image of the point (3, 8) with respect to the line x + 3y = 7, assuming the line to be a plane mirror. ▾

For line x + 3y − 7 = 0, compute d = (3 + 3·8 − 7)/(1² + 3²) = 20/10 = 2.

Image = (3 − 2·1·2, 8 − 2·3·2) = (−1, −4).

Q.9 Find the equation of a line passing through the point of intersection of the lines 4x + 7y − 3 = 0 and 2x − 3y + 1 = 0 that has equal intercepts on the axes. ▾

Solving the two lines gives the intersection (1/13, 5/13).

Equal intercepts ⇒ x + y = a. Through (1/13, 5/13): a = 6/13, so x + y = 6/13 ⇒ 13x + 13y − 6 = 0.

Q.10 A person standing at the junction of two straight paths 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 seeks to reach the path 6x − 7y + 8 = 0 in the least time. Find the equation of the path he should follow. ▾

The least-time path is the perpendicular from the junction to the line 6x − 7y + 8 = 0.

Junction of 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 is (−1/17, 22/17). The required line has slope −7/6 (perpendicular to slope 6/7).

Through (−1/17, 22/17): 119x + 102y − 125 = 0.

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