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

Asked on by cornichon

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

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