Question

Two sides of a parallelogram are along the lines $4x+5y=0$ and $7x+2y=0$. If the equation of oneof the diagonals of the parallelogram is $11x+7y=9$, then other diagonal passes throughthe point :

• A. $(1,2)$
• B. $(2,2)$
• C. $(2,1)$
• D. $(1,3)$

Answer 1

Answer: (B)

Both the lines pass through origin

point $D$ is equal of intersection of $4x+5y=0\&11x+7y=9$
So, coordinates of point $D=\left(\frac{5}{3},-\frac{4}{3}\right)$
Also, point $B$ is point of intersection of $7x+2y=0\&11x+7y=9$
So, coordinates of point $B=\left(-\frac{2}{3},\frac{7}{3}\right)$
diagonals of parallelogram intersect at middle let middle point of $B,D$ $\Rightarrow \left(\frac{\frac{5}{3}-\frac{2}{3}}{2},\frac{-4}{3}+\frac{7}{3}\right)=\left(\frac{1}{2},\frac{1}{2}\right)$
equation of diagonal $AC$
$\Rightarrow (y-0)=\frac{\frac{1}{\alpha }-0}{\frac{1}{\alpha }-0}(\pi -0)$
$y=x$
diagonal AC passes through $(2,2)$