find the values of k so that the area of the triangle with vertices (1,-1),(-4,2k),(-k,-5) is 24 sqaure units.
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let x1 ,y1 =(1,-1)
x2,y2=(-4,2k)
x3,y3=(-k,-5)
now,
24=1/2[2k+20+k-(-5-2k^2+4)
48=3k+20+5+2k^2-4
48=2k^2+3k+21
2k^2+3k-27=0
D=9-4 (2)(-27)
=9+216
=225
k=-3+15/4
k=-3-15/4
= 12/4=3
k=-9/2
x2,y2=(-4,2k)
x3,y3=(-k,-5)
now,
24=1/2[2k+20+k-(-5-2k^2+4)
48=3k+20+5+2k^2-4
48=2k^2+3k+21
2k^2+3k-27=0
D=9-4 (2)(-27)
=9+216
=225
k=-3+15/4
k=-3-15/4
= 12/4=3
k=-9/2
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