Question

Determine algebrically the vertex of the triangle formed by lines $3x-y=3,2x-3y=2,2x-2y=8$

Answer 1

$3x-y=3$ ....(1)

$2x-3y=2$ ...(2)

$2x-2y=8$...(3)

on solving (1) & (2)

We get, $x=1$,$y=0$

on solving (3) & (2)

We get, $x=10$,$y=6$

on solving (1) & (3)

We get, $x=\frac{-1}{2}$,$y=\frac{-9}{2}$

So, vertices of triangle are

$(1,0),(10,6),(\frac{-1}{2},\frac{-9}{2})$