Determinants(Ex. 4.3)

Exercise 4.3

1. Find area of the triangle with vertices at the point given in each of the following:

(i) (1,0), (6,0), (4,3)

Solun:- We know the area of the triangle is:

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

Solun:- We know the area of the triangle is:

(iii) (-2,-3), (3,2), (-1,-8)

Solun:- We know the area of the triangle is:

But We know the area of the triangle is always positive.

Hence Area of Triangle is 15 square units.

2. Show that points

A(a,b+c), B(b,c+a), C(c,a+b) are collinear.

Solun:- We know that if the area of the triangle is zero, then points are collinear.

Area of the triangle ABC is:-

So, the given points are collinear.

3. Find values of k if area of triangle is 4 sq. units and vertices are

(i) (k,0), (4,0), (0,2)

Solun:- We know the area of the triangle is:

Given area of the triangle is 4 sq. units.

⇒ 4 = 1/2 [k(0-2)-0(4-0)+1(8-0)]

⇒ 8 = -2k+8

⇒ k=0

Vertices are (0,0), (4,0), (0,2)

and area = -4

⇒ -4 = 1/2 [k(0-2)-0(4-0)+1(8-0)]

⇒ -8 = -2k+8

⇒ k=8

Vertices are (8,0), (4,0), (0,2)

(ii) (-2,0), (0,4), (0,k)

Solun:- We know area of triangle is:

Given area of the triangle is 4 sq. units.

⇒ 4 = 1/2 [-2(4-k)-0(0-0)+1(0-0)]

⇒ 8 = -8+2k

⇒ k=8

Vertices are (-2,0), (0,4), (0,8)

and area = -4

⇒ -4 = 1/2 [k(0-2)-0(4-0)+1(8-0)]

⇒ -8 = -8+2k

⇒ k=0

Vertices are (-2,0), (0,4), (0,0)

4. (i) Find equation of line joining (1,2) and (3,6) using determinants.

Solun:- Let P(x,y) be a point that lies on the line we know that if three points are collinear then the area of the triangle is zero.

Points are: P(x,y), B(1,2), C(3,6)

For Collinear points the area of the triangle is zero.

⇒ 0 = 1/2 [x(2-6)-y(1-3)+1(0)]

⇒ 0 = -4x+2y

⇒ 4x = 2y

⇒ y = 2x

(ii) Find equation of line joining (3,1) and (9,3) using determinants.

Solun:- Let P(x,y) be a point that lies on the line we know that if three points are collinear then the area of the triangle is zero. Points are: P(x,y), B(3,1), C(9,3)

For Collinear points the area of the triangle is zero.

⇒ 0 = x(1-3)-y(3-9)+1(9-9)

⇒ 0 = -2x+6y

⇒ 0 = -x+3y

⇒ x-3y = 0

5. If the area of the triangle is 35 sq. units with vertices (2,-6), (5,4), (k,4). Then k is

(A) 12     (B) -2     (C) -12, -2     (D) 12,-2

Solun:- We know area of triangle is:

Given area of the triangle is 35 sq. units.

⇒ 35 = 1/2 [2(4-4)-(-6)(5-k)+1(20-4k)]

⇒ 70 = 30-6k+20-4k

⇒ -10k = 20

⇒ k=-2

Vertices are (2,-6), (5,4), (-2,4).

Given area of the triangle is -35 sq. units.

⇒ -35 = 1/2 [2(4-4)-(-6)(5-k)+1(20-4k)]

⇒ -70 = 30-6k+20-4k

⇒ -10k = -120

⇒ k=12

Vertices are (2,-6), (5,4), (12,4).

Answer is…………D.

Download pdf of Exercise 4.3

See also:- Exercise 4.1

Notes of Determinants

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