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Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.

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#### Solution

Take (x_{1},y_{1})=(1,-1),(4,-2k)and (-k,-5)

It is given that the area of the triangle is 24 sq.unit Area of the triangle having vertices (x_{1},y_{1}),(x_{2},y_{2}) and (x_{3},y_{3}) is given by

`=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`

`24=1/2[1(2k-(-5)+(-4)((-5)-(-1))+(-k)((-1)-2k)]`

`48=[(2k+5)+16+(k+2k^2)]`

`2k^2+3k-27=0`

`(2k+9)(k-3)=0`

`k=-9/2 or k=3`

The values of k are -9/2 and 3.

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