question_answer

The area of the triangle formed by the lines \[y=ax,\]\[x+y-a=0\]and the y-axis is equal to

A)  \[\frac{1}{2\left| 1+a \right|}\]

B)  \[\frac{{{a}^{2}}}{\left| 1+a \right|}\]

C)  \[\frac{1}{2}\left| \frac{a}{1+a} \right|\]

D)  \[\frac{{{a}^{2}}}{2\left| 1+a \right|}\]

Correct Answer: D

Solution :

\[B\equiv (0,a),\]\[{{A}_{1}}\equiv \left( \frac{a}{1+a},\frac{{{a}^{2}}}{1+a} \right)\] \[\Delta O{{A}_{1}}B=\frac{1}{2}(OB)\,\,{{A}_{1}}M=\frac{1}{2}\left| a \right|\left| \frac{a}{1+a} \right|=\frac{1}{2}\frac{{{a}^{2}}}{\left| 1+a \right|}\]