Class 10 RD Sharma Solutions - Chapter 14 Coordinate Geometry - Exercise 14.5 | Set 2
Last Updated :
16 May, 2021
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
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