question_answer

One side of length 3a of a triangle of area\[{{a}^{2}}\]square unit lies on the line\[x=a\]. Then, one of the lines on which the third vertex lies, is

A)  \[x=-{{a}^{2}}\]                               

B)  \[x={{a}^{2}}\]

C)  \[x=-a\]              

D)         \[x=\frac{a}{3}\]

E)  \[x=-\frac{a}{3}\]

Correct Answer: D

Solution :

Area of\[\Delta ABC={{a}^{2}}\] \[\Rightarrow \]               \[\frac{1}{2}(a-x)3a={{a}^{2}}\]                      \[\Rightarrow \]               \[a-x=\frac{2}{3}a\] \[\Rightarrow \]               \[x=\frac{a}{3}\] Hence, one of the line on which third vertex lies is\[x=\frac{a}{3}\].