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

• Last Updated : 08 May, 2021

### (i) (3, 8), (âˆ’4, 2) and (5, âˆ’1)

Solution:

Given (3, 8), (âˆ’4, 2) and (5, âˆ’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) (âˆ’1, âˆ’8), (âˆ’2, âˆ’3) and (3, 2)

Solution:

Given (âˆ’1, âˆ’8), (âˆ’2, âˆ’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), (âˆ’5, 1) and (10, 7)

Solution:

Given points are (5, 5), (âˆ’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, âˆ’1), (2, 1) and (4, 5)

Solution:

Given points are (1, âˆ’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, âˆ’2), (8, 8) and (5, 2)

Solution:

Given points are (3, âˆ’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), (âˆ’1, âˆ’2) and (5, 8)

Solution:

Given points are (2, 3), (âˆ’1, âˆ’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 âˆ’ a âˆ’ b = 0

=> a + b = ab

Hence proved.

### Question 4. Using the determinants prove that the points (a, b), (aâ€™, bâ€™), and (a â€“ aâ€™, b â€“ b) are collinear if a bâ€™ = aâ€™ b.

Solution:

Given points are (a, b), (aâ€™, bâ€™) and (a â€“ aâ€™, b â€“ b) and a bâ€™ = aâ€™ b.

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

=>

=>

=>

=> a bâ€™ âˆ’ aâ€™ b = 0

=> a bâ€™ = aâ€™ b

Hence proved.

### Question 5. Find the value of Î» so that the points (1, âˆ’5), (âˆ’4, 5), and (Î», 7) are collinear.

Solution:

Given points are (1, âˆ’5), (âˆ’4, 5) and (Î», 7).

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

=>

=>

=> âˆ’2 âˆ’ 20 âˆ’ 5Î» âˆ’ 28 âˆ’ 5Î» = 0

=> 10Î» = 50

=> Î» = 5

Therefore, the value of Î» is 5.

### Question 6. Find the value of x if the area of âˆ† is 35 square cms with vertices (x, 4), (2, âˆ’6), and (5, 4).

Solution:

Given points are (x, 4), (2, âˆ’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 =

=>

=>

=> âˆ’ 10x + 12 + 38 = Â±70

=> â€“ 10x + 50 = Â±70

Taking positive sign, we get

=> â€“ 10x + 50 = 70

=> 10x = â€“ 20

=> x = â€“ 2

Taking negative sign, we get

=> â€“ 10x + 50 = â€“ 70

=> 10x = 120

=> x = 12

Therefore, the value of x is 12 or â€“2.

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

Solution:

Given points are (1, 4), (2, 3), (â€“5, â€“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 (â€“3, 5), (3, â€“6), (7, 2).

Given points are (â€“3, 5), (3, â€“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, 2â€“2k), (â€“k+1, 2k), (â€“4â€“k, 6â€“2k) may be collinear.

Solution:

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

=>

=> k(2k â€“ 6 + 2k) â€“ (2â€“2k) (â€“k + 1 + 4 + k) + [(1â€“k) (6â€“2k) â€“ 2k (â€“4â€“k)] = 0

=> 8k2 + 4k â€“ 4 = 0

=> 8k2 + 8k â€“ 4k â€“ 4 = 0

=> 8k (k+1) â€“ 4 (k+1) = 0

=> 8k = 4 or k = â€“1

=> k = 1/2 or k = â€“1

Therefore, the value of k is 1/2 or â€“1.

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

Solution:

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

=>

=> x(2 â€“ 8) + 2(5 â€“ 8) + 1(40â€“16) = 0

=> â€“6x â€“ 6 + 24 = 0

=> 6x = 18

=> x = 3

Therefore, the value of x is 3.

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

Solution:

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

=>

=> 3(2 â€“ 8) + 2(x â€“ 8) + 1(8xâ€“16) = 0

=> â€“18 + 2x â€“ 16 + 8x â€“ 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,

=>

=>

=> âˆ’4x + 2y = 0

=> 2x âˆ’ y = 0

Therefore, the required equation is 2x âˆ’ 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,

=>

=>

=> âˆ’2x + 6y = 0

=> x âˆ’ 3y = 0

Therefore, the required equation is x âˆ’ 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(0âˆ’2) âˆ’ 0 + 1(8âˆ’0)| = 8

=> âˆ’2k + 8 = Â±8

=> âˆ’2k + 8 = 8 or âˆ’2k + 8 = âˆ’8

=> k = 0 or k = 8

Therefore, the value of k is 0 or 8.

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

Solution:

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

=>

=> |âˆ’2(4âˆ’k) âˆ’ 0 + 1(0)| = 8

=> âˆ’8 + 2k = Â±8

=> âˆ’8 + 2k = 8 or âˆ’8 + 2k = âˆ’8

=> k = 8 or k = 0

Therefore, the value of k is 0 or 8.

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