Class 10 RD Sharma Solutions – Chapter 14 Coordinate Geometry – Exercise 14.5 | Set 2

Question 12. Prove that the points (a, b), (a1, b1) and (a – a1, b – b1) are collinear if ab1 = a1b.

Solution:

As we know that 3 points are collinear is the area of the triangle formed by them is zero

So, let us assume ABC is a triangle whose vertices A(a, b), B(a1, b1) and C(a – a1, b – b1)

Area of triangle = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

= 1/2 [a(b1 – b + b1) + a1(b – b1 – b) + (a – a1)(b – b1)]

= 1/2 [2b1a – b – a1b1 + ab – ab1 – a1b + a1b1]

= 1/2 (ab1 – a1b)

The points are collinear

So, Area of ∆ABC = 0

1/2(ab1 – a1b) = 0    

⇒ ab1 – a1b = 0

⇒ ab1 = a1b

Hence, Proved

Question 13. If the vertices of a triangle are (1, -3), (4, p), and (-9, 7) and its area is 15 square units, find the value(s) of p. 

Solution:

Let us assume ABC is a triangle whose vertices are A(1, -3), B(4, p), and C(-9, 7)

It is given that, area of triangle = 15 sq. units

So, Area of ∆ABC = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

15 = 1/2[1(p – 7) + 4(7 + 3) + (-9)(-3 – p)]

15 = 1/2[p – 7 + 40 + 27 + 9p]

15 = 1/2[10p + 60] = 5p + 30

5p + 30 = 15

5p = 15 – 30 = -15

p = -15/5 = -3

Hence, the value of p is -3

Question 14. If (x, y) be on the line joining the two points (1, -3) and (-4, 2), prove that x + y + 2 = 0.

Solution:

Given that point (x, y) be on the line joining the two points (1, -3) and (-4, 2)

So, points (x, y), (1, -3) and (-4, 2) are collinear

Now, let us assume that the points (x, y) (1, -3) and (-4, 2) are the vertices of a triangle ABC, 

So, Area of ∆ABC = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

= 1/2[x(-3 – 2) + 1(2 – y) + (-4)(y + 3)]

= 1/2[-5x + 2 – y – 4y – 12]

= 1/2[-5x – 5y – 10] 

= -5/2(x + y + 2)

As we know that the points are collinear 

So, the area of ∆ABC = 0

⇒ =5/2 (x + y + 2) = 0

⇒ x + y + 2 = 0

Question 15. Find the value of k if points (k, 3), (6, -2), and (-3, 4) are collinear. 

Solution:

Let us assume ABC is a triangle whose vertices are A(k, 3), B(6, -2), and C(-3, 4)

So, 

Area of ∆ABC = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

= 1/2[k(-2 – 4) + 6(4 – 3) + (-3)(3 + 2)]

= 1/2[-6k + 6 * 1 + (-3 * 5)]

= 1/2[-6k + 6 – 15]

= 1/2[-6k – 9]

= -3/2 [2k + 3]

It is given that the points are collinear 

So, the area of ∆ABC = 0

-3/2 [2k + 3] = 0

 ⇒2k = -3 ⇒ k = -3/2

Hence, the value of k is-3/2

Question 16. Find the value of k, if the points A (7, -2), B (5, 1), and C (3, 2k) are collinear.

Solution:

Let us assume ABC is a triangle whose vertices are A (7, -2), B (5, 1), and C (3, 2k) 

So, 

Area of ∆ABC = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

= 1/2 [7(1 – 2k) + 5(2k + 2) + 3(-2 – 1)]

= 1/2 [7 – 14k + 10k + 10 – 6 – 3]

= 1/2 [8 – 4k] = 4 – 2k

It is given that the points are collinear 

So, the area of ∆ABC = 0

⇒ 4 – 2k = 0

⇒ 2k = 4

⇒ k = 2

Hence, the value of k is 2

Question 17. If the point P (m, 3) lies on the line segment joining the points A (−2/5, 6) and B (2, 8), find the value of m.

Solution:

It is given that points P(m, 3) lies on the line segment joining the points A (−2/5, 6) and B (2, 8)

So, points A, P, B are collinear

Now the area of area ∆APB = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

= 1/2[-2/5(3 – 8) + m(8 – 6) + 2(6 – 3)]

= 1/2[-2/5 * (-5) + 2m + 2 * 3]

= 1/2[2 + 2m + 6]

= 1/2 [2m + 8] 

= m + 4

As we know that the points are collinear 

So, the area of ∆APB = 0

m + 4 = 0

⇒m = -4

Hence, the value of m is -4

Question 18. If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that x + y = a + b. 

Solution:

It is given that point R (x, y) lies on the line segment joining the points P (a, b) and Q (b, a)

So, points R, P, Q are collinear

Now area of ∆PRQ = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

= 1/2 [a(y – a) + x(a – b) + b(b – y)]

= 1/2 [ay – a² + ax – bx + b² – by] 

= 1/2 [x(a – b) + y(a – b) – (a² – b²)]

= 1/2 [x(a – b) + y(a – b) – (a + b)(a – b)]

= 1/2 (a – b)(x + y – a – b) = 0

⇒x + y – a – b = 0

⇒x + y = a + b

Hence, c + y = a + b

Question 19. Find the value of k, if the points A (8, 1), B (3, -4), and C (2, k) are collinear.

Solution:

Let us assume ABC is a triangle whose vertices are A (8, 1), B (3, -4), and C (2, k)

So, 

Area of ∆ABC = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

= 1/2[8(-4 – k) + 3(k – 1) + 2(1 + 4)]

= 1/2[-32 – 8k + 3k – 3 + 10] 

= 1/2 [-25 – 5k]

It is given that the points are collinear 

So, the area of ∆ABC = 0

⇒ 1/2[-25 -5k] = 0

⇒ 1/2 × (-5)(5 + k) = 0

⇒ 5 + k = 0

⇒ k = -5

Hence, the value of m is -5

Question 20. Find the value of a for which the area of the triangle formed by the points A (a, 2a), B (-2, 6), and C (3, 1) is 10 square units.

Solution:

It is given that, ABC is a triangle whose vertices are A(a, 2a), B(-2, 6), and C(3, 1)

So, Area of ∆ABC = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

= 1/2[a(6 – 1) + (-2)(1 – 2a) + 3(2a – 6)] 

= 1/2[5a – 2 + 4a + 6a – 18] 

= 1/2[15a – 20]

It is given that the area of ∆ABC = 10 sq. units

So, 

1/2[15a – 20] = 10

⇒ 15a – 20 = 20

⇒ 15a = 20 + 20 = 40

⇒a = 40/15 = 8/3

Hence, the value of a is 8/3

Question 21. If a ≠ b ≠ 0, prove that the points (a, a²), (b, b²), (0, 0) are never collinear. 

Solution:

Let us assume that A(a, a²), B(b, b²), and C(0, 0) be the coordinates of the given points.

As we know that the area of a triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

= 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)] 

So area of ∆ABC = 1/2[a(b² – 0) + b(0 – a²) + 0(a² – b²)]

= 1/2[ab² – a²b]

= 1/2[ab(b – a)] ≠ 0                             

Therefore, a ≠ b ≠ 0

Hence, the area of the triangle formed by the points (a, a²), (b, b²), (0, 0) is not zero, 

so the given points are not collinear

Question 22. The area of a triangle is 5 sq. units. Two of its vertices are at (2, 1) and (3, -2). If the third vertex is (7/2, y), find y.

Solution:

Let us assume ABC is a triangle whose vertices are A(2, 1), B(3, -2), and C (7/2, y)

Also, the area of ∆ABC = 5 sq.units

So, 

Area of ∆ABC = 1/2[x₁(y₂ – y₃) + (y₃ – y₁)x₂ + x₃(y₁ – y₂)]

⇒ ±5 = 1/2[2(-2 – y) + 3(y – 1) + 7/2(1 + 2)] 

⇒ ±10 = [-4 – 2y + 3y – 3 + 21/2]

⇒ ±10 = [y – 7 + 21/2]

⇒ ±10 = (2y – 14 + 21)/2

⇒ 2y + 7 = ±20

⇒ 2y = 20 – 7 or 2y = -20 – 7

⇒ y = 13/2 or y = -27/2