Answer:
The vertices of the given \[\Delta \text{ }ABC\] are \[A(1,-1),B(-4,2k)\] and \[C(-k,-5)\] \[\therefore \] Area of \[\Delta \,ABC=\frac{1}{2}[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]\] \[=\frac{1}{2}[1(2k+5)+(-4)(-5+1)+(-k)(-1-2k)]\] \[=\frac{1}{2}[2k+5+16+k+2{{k}^{2}}]\] \[=\frac{1}{2}[2{{k}^{2}}+3k+21]\] Area of \[\Delta \,ABC=24\]sq. units (Given) \[\therefore \frac{1}{2}[2{{k}^{2}}+3k+21]=24\] \[[2{{k}^{2}}+3k+21]=48\] \[\Rightarrow 2{{k}^{2}}+3k+21=48\] \[\Rightarrow 2{{k}^{2}}+3k-27=0\] \[\Rightarrow 2{{k}^{2}}+9k-6k-27=0\] \[\Rightarrow k(2k+9)-3(2k+9)=0\] \[\Rightarrow (k-3)(2k+9)=0\] \[k=3\]or \[k=-\frac{9}{2}\] Hence. \[k=3\] or \[k=-\frac{9}{2}\].
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