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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