# Find the area of the triangle formed by the lines: 1) 11x-7y=81 2) 3x-5y=-15 3) x+4y=12

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We have the three lines which form the sides of the triangle as

11x-7y=81...(1)

3x-5y=-15...(2)

x+4y=12...(3)

We have to find their points of intersection.

11*(2) - 3*(1)

=> 33x - 55y - 33x + 21y = -408

=> -34y = 408

=> y = 12

x = (81 + 7*12)/11

= 15

**So one vertex is (15,12)**

(2) - 3*(3)

=> 3x-5y- 3x - 12y = -15 - 36

=> -17y = -51

y = 3

x = 12 - 4*3 = 0

**The other vertex is ( 0,3)**

(1) - 11*(3)

=> 11x-7y - 11x - 44y =81 - 132

=> -51y = -51

=> y = 1

x= 12 - 4*1 = 8

**The third vertex is (8 , 1)**

Now the distance between (8,1) and (0,3) = sqrt(8^2+2^2) = sqrt(68)

The equation of the line between these points is

y - 3 = (1-3)/(8-0)*x

=> y - 3 = -2x/8

=> 8y - 24 = -2x

=> 2x + 8y - 24 = 0

The perpendicular distance between (15, 12) and 2x + 8y - 24 = 0 is

| 2*15 + 8*12 - 24|/ sqrt ( 2^2 + 8^2)

=> 30 +96 - 24/sqrt( 4+ 64)

=> 102 / sqrt 68

The area of the triangle can be calculated by the formula (1/2)*base*height = (1/2)*sqrt(68)*(102 / sqrt 68)

=> 102/2

=> 51

**Therefore the required area is 51.**

### 3) x+4y=12

We solve the system of equations of the lines and then solve for the area of the triangle formed by the points of intersection of the lines.

1) 11x-7y=81.

2) 3x-5y=-15.

3) x+4y=12.

(1)3 -(2)*11 : (-7*3+5*11)y = 81*3+15*11= 408, or 34y = 408. so y = 408/34 = 12. Putting y = 12 in (2), we get 3x = -15+5y = -15+5*12 = 45. So x = 45/3 = 15, So intersection of the lines (1) and (2) is at **(15, 12)**

(3)*11- (1): (4*11+7)y = 12*11-81 = 51, or 51y = 51,,So y = 1. Substituting y = 1 in (3), we get x+4 = 12, so x= 12-4 = 8. So intersection of the lines (1) and (3) is at **(8,1**)

5)y = 12*3-(3)*3 - (2) gives (4*3+5)y = 12*3+15 = 51, 1y = 51, so y = 51/17 = 3. Put y = 3 in (3), x+4*3 = 12, so x= 0. So the intersection of the lines (2) and (3) is at **(0, 3).**

So the intersection of the 3 lines at (15,12), (8,1) and (0, 3) form the vertices of the triangle.

Let (x1,y1) = (0,3), (x2,y2) = (8,1) and (x3,y3) = (15,12).

Then the area A of the triangle is given by:

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

A= (1/2)|{(8-0)(3+1)+(15-8)(12+1)+(0-15)(3+12)}|.

A= (1/2){8*4+7*13-15*15}|.

A = (1/2)|{32+91 - 225}|.

A = (1/2)(102).

A = 51 sq units.

Therefore the area of the triangle is 51 sq units.