A) -2
B) -4
C) - 6
D) 8
Correct Answer: C
Solution :
Given that \[{{x}_{1}}=x,\,\,{{x}_{2}}=1,\,{{x}_{3}}=0\] \[\therefore \] \[{{y}_{1}}=0,\,{{y}_{2}}=1,\,{{y}_{3}}=2\] \[\therefore \] Area of triangle \[=\frac{1}{2}\,[{{x}_{1}}\,({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]\] \[=\frac{1}{2}\,[x\,(1-2)+1(2-0)+(0-1)]\] \[=\frac{1}{2}\,[-x\,+2+0]\] \[=\frac{1}{2}\,(2-x)\] Now, \[\frac{1}{2}(2-x)=4\] sq. units (given) \[\Rightarrow \] \[2-x=8\] \[\Rightarrow \] \[x=-6\]You need to login to perform this action.
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