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

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We have the three lines 4x – y – 9 = 0 , 3x – 2y – 3 = 0 and x – y = 0.

First, let us find their points of intersection.

We can write 4x – y – 9 = 0 as y = 4x – 9

Substitute this in 3x – 2y – 3 = 0

=> 3x – 2(4x – 9) – 3 =0

=> 3x – 8x + 18 – 3 = 0

=> -5x = -15

=> x = 3

y = 4x – 9 = 3

Now we see that (3 , 3) also lies on x – y = 0

Therefore all the three lines intersect at the same point. So they do not form a triangle.

**The three lines do not form a triangle.**

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.**