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

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