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 easily understandable language, which improves the grasping abilities of students. The presentation of each solution in the chapter is done in a unique way by BYJU’S experts in Maths.
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RD Sharma Solutions for Class 12 Chapter 6 Determinants Exercise 6.3
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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 units is the area
Thus the area of the 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 if three points are 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 if three points are collinear, the area of the triangle formed by these points will be zero. Now, we know that, the 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 if three points are 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 if three points are 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 if three points are 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 if three points are 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 if three points are 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 if three points are 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
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