 GeeksforGeeks App
Open App Browser
Continue

## Related Articles

• RD Sharma Class 12 Solutions for Maths

# Class 12 RD Sharma Solutions – Chapter 6 Determinants – Exercise 6.3

### (i) (3, 8), (−4, 2) and (5, −1)

Solution:

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,

A =     Therefore, area of the triangle is sq. units.

### (ii) (2, 7), (1, 1) and (10, 8)

Solution:

Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle.

The area of the triangle is given by,

A =    Therefore, area of the triangle is sq. units.

### (iii) (−1, −8), (−2, −3) and (3, 2)

Solution:

Given (−1, −8), (−2, −3) and (3, 2) are the vertices of the triangle.

The area of the triangle is given by,

A =    = 15

Therefore, area of the triangle is 15 sq. units.

### (iv) (0, 0), (6, 0), (4, 3)

Solution:

Given (0, 0), (6, 0), (4, 3) are the vertices of the triangle.

The area of the triangle is given by,

A =   = 9

Therefore, area of the triangle is 9 sq. units.

### (i) (5, 5), (−5, 1) and (10, 7)

Solution:

Given points are (5, 5), (−5, 1) and (10, 7).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

A =   = 0

As the area of the triangle is 0, the points are collinear.

Hence proved.

### (ii) (1, −1), (2, 1) and (4, 5)

Solution:

Given points are (1, −1), (2, 1) and (10, 8).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

A =   = 0

As the area of the triangle is 0, the points are collinear.

Hence proved.

### (iii) (3, −2), (8, 8) and (5, 2)

Solution:

Given points are (3, −2), (8, 8) and (5, 2).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

A =   = 0

As the area of the triangle is 0, the points are collinear.

Hence proved.

### (iv) (2, 3), (−1, −2) and (5, 8)

Solution:

Given points are (2, 3), (−1, −2) and (5, 8).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

A =   = 0

As the area of the triangle is 0, the points are collinear.

Hence proved.

### Question 3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.

Solution:

Given points (a, 0), (0, b) and (1, 1) are collinear.

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

=> => => ab − a − b = 0

=> a + b = ab

Hence proved.

### Question 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 points are (a, b), (a’, b’) and (a – a’, b – b) and a b’ = a’ b.

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

=> => => => a b’ − a’ b = 0

=> a b’ = a’ b

Hence proved.

### Question 5. Find the value of λ so that the points (1, −5), (−4, 5), and (λ, 7) are collinear.

Solution:

Given points are (1, −5), (−4, 5) and (λ, 7).

For these points to be collinear, the area of the triangle they form must be zero. So, we get,

=> => => −2 − 20 − 5λ − 28 − 5λ = 0

=> 10λ = 50

=> λ = 5

Therefore, the value of λ is 5.

### Question 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 points are (x, 4), (2, −6) and (5, 4).

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,

A = => => => − 10x + 12 + 38 = ±70

=> – 10x + 50 = ±70

Taking positive sign, we get

=> – 10x + 50 = 70

=> 10x = – 20

=> x = – 2

Taking negative sign, we get

=> – 10x + 50 = – 70

=> 10x = 120

=> x = 12

Therefore, the value of x is 12 or –2.

### Question 7. Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3), (–5, –3). Are the given points collinear?

Solution:

Given points are (1, 4), (2, 3), (–5, –3).

So, area =    As the area of the triangle formed by these three points is not zero, the points are not collinear.

### Question 8. Using determinants, find the area of the triangle with vertices (–3, 5), (3, –6), (7, 2).

Given points are (–3, 5), (3, –6), (7, 2).

So, area =    = 46

Therefore, the area of the triangle is 46 sq. units.

### Question 9. Using determinants, find the value of k so that the points (k, 2–2k), (–k+1, 2k), (–4–k, 6–2k) may be collinear.

Solution:

Given points are (k, 2–2k), (–k+1, 2k), (–4–k, 6–2k). As the points are collinear, the area must be 0.

=> => k(2k – 6 + 2k) – (2–2k) (–k + 1 + 4 + k) + [(1–k) (6–2k) – 2k (–4–k)] = 0

=> 8k2 + 4k – 4 = 0

=> 8k2 + 8k – 4k – 4 = 0

=> 8k (k+1) – 4 (k+1) = 0

=> 8k = 4 or k = –1

=> k = 1/2 or k = –1

Therefore, the value of k is 1/2 or –1.

### Question 10. If the points (x, –2), (5, 2), (8, 8) are collinear, find x using determinants.

Solution:

Given points are (x, –2), (5, 2), (8, 8). As the points are collinear, the area must be 0.

=> => x(2 – 8) + 2(5 – 8) + 1(40–16) = 0

=> –6x – 6 + 24 = 0

=> 6x = 18

=> x = 3

Therefore, the value of x is 3.

### Question 11. If the points (3, –2), (x, 2), (8, 8) are collinear, find x using determinants.

Solution:

Given points are (3, –2), (x, 2), (8, 8). As the points are collinear, the area must be 0.

=> => 3(2 – 8) + 2(x – 8) + 1(8x–16) = 0

=> –18 + 2x – 16 + 8x – 16 = 0

=> 10x = 50

=> x = 5

Therefore, the value of x is 5.

### (i) the line joining the points (1, 2) and (3, 6).

Solution:

Let (x, y), (1, 2), (3, 6) be the points on the line. As these points are collinear, we get,

=> => => −4x + 2y = 0

=> 2x − y = 0

Therefore, the required equation is 2x − y = 0.

### (ii) the line joining the points (3, 1) and (9, 3).

Solution:

Let (x, y), (3, 1), (9, 3) be the points on the line. As these points are collinear, we get,

=> => => −2x + 6y = 0

=> x − 3y = 0

Therefore, the required equation is x − 3y = 0.

### (i) if area of triangle whose vertices are (k, 0) (4, 0) (0, 2) is 4 square units.

Solution:

Given points are (k, 0) (4, 0) (0, 2). According to the question, we have,

=> => |k(0−2) − 0 + 1(8−0)| = 8

=> −2k + 8 = ±8

=> −2k + 8 = 8 or −2k + 8 = −8

=> k = 0 or k = 8

Therefore, the value of k is 0 or 8.

### (ii) if area of triangle whose vertices are (−2, 0) (0, 4) (0, k) is 4 square units.

Solution:

Given points are (−2, 0) (0, 4) (0, k). According to the question, we have,

=> => |−2(4−k) − 0 + 1(0)| = 8

=> −8 + 2k = ±8

=> −8 + 2k = 8 or −8 + 2k = −8

=> k = 8 or k = 0

Therefore, the value of k is 0 or 8.

My Personal Notes arrow_drop_up
Related Tutorials