question_answer

If the coordinates of the points A, B, C, D, be \[(a,\ b),\] \[({a}',\ {b}'),\] \[(-a,\ b)\] and \[({a}',\ -{b}')\] respectively, then the equation of the line bisecting the line segments AB and CD is

A)            \[2{a}'y-2bx=ab-{a}'{b}'\]    

B)            \[2ay-2{b}'\ x=ab-{a}'{b}'\]

C)            \[2ay-2{b}'x={a}'b-a{b}'\]    

D)            None of these

Correct Answer: B

Solution :

           Mid point of \[AB=E\text{ }\left( \frac{a+{a}'}{2},\frac{b+{b}'}{2} \right)\]and mid point of \[CD=F\text{ }\left( \frac{{a}'-a}{2},\frac{b-{b}'}{2} \right)\]. Hence equation of line EF is                    \[y-\frac{b+{b}'}{2}\]\[=\frac{b-{b}'-b-{b}'}{{a}'-a-a-a'}\left( x-\frac{a+{a}'}{2} \right)\]                    On simplification, we get \[2ay-2{b}'x-=ab-{a}'{b}'\].