RD Sharma Solutions for Class 12 Maths Exercise 6.3 Chapter 6 Determinants is provided here. The third exercise of Chapter 6 explains the applications of determinants to coordinate geometry. The solutions are explained in understandable language which improves grasping abilities among students. The presentation of each solution in the chapter is described in a unique way by the BYJU’S experts in Maths.
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Exercise 6.3 Page No: 6.71
1. Find the area of the triangle with vertices at the points:
(i) (3, 8), (-4, 2) and (5, -1)
(ii) (2, 7), (1, 1) and (10, 8)
(iii) (-1, -8), (-2, -3) and (3, 2)
(iv) (0, 0), (6, 0) and (4, 3)
Solution:
(i) Given (3, 8), (-4, 2) and (5, -1) are the vertices of the triangle.
We know that, if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by:
(ii) Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle.
We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by:
(iii) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.
We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by:
As we know area cannot be negative. Therefore, 15 square unit is the area
Thus area of triangle is 15 square units
(iv) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.
We know that if vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by:
2. Using the determinants show that the following points are collinear:
(i) (5, 5), (-5, 1) and (10, 7)
(ii) (1, -1), (2, 1) and (10, 8)
(iii) (3, -2), (8, 8) and (5, 2)
(iv) (2, 3), (-1, -2) and (5, 8)
Solution:
(i) Given (5, 5), (-5, 1) and (10, 7)
We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by
(ii) Given (1, -1), (2, 1) and (10, 8)
We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,
(iii) Given (3, -2), (8, 8) and (5, 2)
We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,
Now, by substituting given value in above formula
Since, Area of triangle is zero
Hence, points are collinear.
(iv) Given (2, 3), (-1, -2) and (5, 8)
We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,
3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab
Solution:
Given (a, 0), (0, b) and (1, 1) are collinear
We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,
⇒
⇒ a + b = ab
Hence Proved
4. Using the determinants prove that the points (a, b), (a’, b’) and (a – a’, b – b) are collinear if a b’ = a’ b.
Solution:
Given (a, b), (a’, b’) and (a – a’, b – b) are collinear
We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,
⇒ a b’ = a’ b
Hence, the proof.
5. Find the value of λ so that the points (1, -5), (-4, 5) and (λ, 7) are collinear.
Solution:
Given (1, -5), (-4, 5) and (λ, 7) are collinear
We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,
⇒ – 50 – 10λ = 0
⇒ λ = – 5
6. Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, -6) and (5, 4).
Solution:
Given (x, 4), (2, -6) and (5, 4) are the vertices of a triangle.
We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1, y1), (x2, y2) and (x3, y3), then the area of the triangle is given by,
⇒ [x (– 10) – 4(– 3) + 1(8 – 30)] = ± 70
⇒ [– 10x + 12 + 38] = ± 70
⇒ ±70 = – 10x + 50
Taking positive sign, we get
⇒ + 70 = – 10x + 50
⇒ 10x = – 20
⇒ x = – 2
Taking –negative sign, we get
⇒ – 70 = – 10x + 50
⇒ 10x = 120
⇒ x = 12
Thus x = – 2, 12
