HOW TO FIND AREA OF TRIANGLE WHEN EQUATION OF SIDES ARE GIVEN

Here, we are going to see an example problem to learn how to find area of triangle when equations of the sides are given.

Question :

Find the area of a triangle formed by the lines

3x + y − 2  =  0

5x + 2y − 3  =  0

2x − y − 3  =  0

Answer :

3x + y − 2 = 0  ------(1)

5x + 2y − 3 = 0  ------(2)

2x − y − 3 = 0  ------(3)

Point of intersection of (1) and (2) is the vertex A

Point of intersection of (2) and (3) is the vertex B

Point of intersection of (3) and (1) is the vertex C

2(1) - (2) : 

x - 1  =  0

Substitute 1 for x in (1).

(1)-----> 3(1) + y - 2  =  0

1 + y  =  0

y  =  -1

Vertex A(1, -1).

(2) + 2(3) :

9x - 9  =  0

9x  =  9

x  =  1

Substitute 1 for x in (2).

(2)-----> 5(1) + 2y - 3  =  0 

5 + 2y - 3  =  0

2y + 2  =  0

2y  =  -2

y  =  -1

Vertex B (1, -1)

(1) + (3) :

5x - 5  =  0

x  =  1

Substitute 1 for x (3).

(1)-----> 2(1) - y - 3  =  0

2 - y - 3  =  0

- y - 1  =  0

-y  =  1

y  =  -1

Vertex C(1, -1). 

Vertices of the triangle are 

A(1, -1) B(1, -1) and C(1, -1)

Now, let us find the area of triangle using the above vertices.

  =  (1/2)[(-3 +3)]

  =  0 square units

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