Question

The sides of the triangle are on the of x+y+6=0, x+2y-4=0, 3x+y+2=0. Find the vertices of the triangle

Answer 1

Given three non-parallel lines.
We are to find the vertices of a triangle formed by the intersection of the lines.

The intersections are found by solving for the three systems of two equations at a time.

For example:
L1: x+y+6=0
L2: x+2y-4=0
Subtract L1 from L2 gives:
y-10=0, this means that
y=10
Now substitute y=10 in equation L1 gives x+10+6=0
x=-16.
So (-16,10) is one of the vertices.

Repeat the same for lines L2 & L3, and then L1 & L3 will give the two other vertices.