A) \[6x+y-32=0\]
B) \[6x-y+32=0\]
C) \[x+6y-32=0\]
D) \[6x+y+32=0\]
Correct Answer: A
Solution :
Given coordinates of a triangle are \[A(1,\,\,5)\] and\[B(3,\,\,-7)\]. Let the coordinate of third vertex of a triangle be\[({{x}_{1}},\,\,{{y}_{1}})\]. \[\therefore \]Area of \[\Delta ABC=21\](given) \[\Rightarrow \]\[\frac{1}{2}[1(-7-{{y}_{1}})+3({{y}_{1}}-5)+{{x}_{1}}(5+7)]=21\] \[\Rightarrow \] \[2{{y}_{1}}+12{{x}_{1}}-22=42\] \[\Rightarrow \] \[12{{x}_{1}}+2{{y}_{1}}-64=0\] \[\Rightarrow \] \[6{{x}_{1}}+{{y}_{1}}-32=0\] \[\therefore \]Locus of a point is \[6x+y-32=0\]
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