Two sides of a parallelogram are along lines 4 x+5 y=0 and 7 x+2 y=0 . If equation of one of diagona

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9.Straight Line

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

A$(1,3)$

B$(1,2)$

C$(2,2)$

D$(2,1)$

(JEE MAIN-2021)

Solution

Both the lines pass through origin.
point $D$ is equal of intersection of $4 x+5 y=0\, \,11 x+7 y=9$
So, coordinates of point $D=\left(\frac{5}{3},-\frac{4}{3}\right)$
Also, point $B$ is point of intersection of $7 x+2 y=0\, \, 11 x+7 y=9$
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 $AC$
$\Rightarrow(y-0)=\frac{\frac{1}{2}-0}{\frac{1}{2}-0}(x-0)$
$y=x$
diagonal $AC$ passes through $(2,2)$

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Solution

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