# 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. - Mathematics

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.

#### Solution

Take (x1,y1)=(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 (x1,y1),(x2,y2) and (x3,y3) 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.

Concept: Area of a Triangle
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