Find the vertices of triangle which sides are on the lines 2x+y-3=0; x-y-6=0; -2x+y-5=0?

To determine the vertices of the triangle whose sides are
along the given lines, we'll have to determine the intercepting points of these
lines.

We'll determine the intercepting point of the lines
2x+y-3=0 and x-y-6=0.

We'll solve the system of equations
using elimination. We'll add the equations:

2x + y - 3 + x
- y - 6 = 0

3x - 9 = 0

3x = 9
=> x = 3

3 - y - 6 = 0 => -y = 6 -
3

y = -3

The first
intercepting point and the 1st vertex  of the triangle is the pair (3 ;
-3).

We'll determine the next intercepting point of the
lines x-y-6=0; -2x+y-5=0.

We'll add the
equations:

x - y - 6 - 2x + y - 5 =
0

-x - 11 = 0 => x =
-11

-11 - y - 6 = 0

y =
-17

The 2nd intercepting point and the 2nd vertex  of the
triangle is the pair (-11 ; -17).

We'll determine the 3d
intercepting point of the lines 2x+y-3=0; -2x+y-5=0.

We'll
add the equations:

2x + y - 3 - 2x + y - 5 =
0

2y - 8 = 0

2y = 8 =>
y = 4

2x + 4 - 3 = 0

2x + 1 =
0

x = -1/2

The 3rd
intercepting point and the 3rd vertex  of the triangle is the pair (-1/2 ;
4).

The vertices of triangle are represented
by the following pairs: (3 ; -3) ; (-11 ; -17) ; (-1/2 ;
4).