What is the area of the triangle formed by the lines 4x – y – 9 = 0 , 3x – 2y – 3 = 0 and x – y = 0

To determine the area of triangle formed by
:

4x-y -9 = 0 (1)

 3x-y-3 = 0
..(2) and

  x-y = 0....
(3):

From (3) , x= y. So from  (2), we get: 3x-x -3 = 0. 2x
= 3, So (x= 3/2. Or (x1, y1) =
(3/2,3/2).

From (3) and (1), we get:  4x-y =
9, or 4x-x = 9, or x= y = 9/3. So (x2, y2) =
(3,3).

From (1) and (2): y = 4x-9. and 3x-
2y = 3, Or 3x- 2(4x-9) = 3. So 3x-8x +18 = 3. So -5x = 3-18 = -15. So x2 = -15/-5 = 3.
This gives y =4*3-9 = 3, or y = 3. Or (x3,y3) = (3,
3).

Therefore the area A of the triangle
formed by the intersections of the lines is are enclosed by the 3 vertices
(x1,y1),(x2,y2) and (x3,y3) which is given by:

A =
(1/2){(x2-x1)(y1+y2)+(x3-x2)(y2+y3)+(x1-x3)(y1+y3)}

A=
(1/2){(3-3/2)(3/2+3)+(3-3)(3+3)+(3/2-3)(3/2+3)}

A=
(1/2){(3/2)3 + 0 -(3/2)3}

A =
0.

Therefore the area of the triangle formed
by the lines is zero.