Question

Two sides of a rhombus are along the lines $x-y+1=0$ and $7x-y-5=0$.
If its diagonals at $(-1,-2)$, then which one of the following is a vertex of this rhombus?

• A. $(-3,-9)$
• B. $(-3,-8)$
• C. $(\frac{1}{3},-\frac{8}{3})$
• D. $(-\frac{10}{3},-\frac{7}{3})$

Answer 1

The correct option is C $(\frac{1}{3},-\frac{8}{3})$

$x-y+1=0$ and $7x-y=5$

Point of intersection

$\to x-y=-1$

$7x-y=5$

$-$

$-6x=-6$

$x=1$

$y=x+1=2$

Slope of AE$=\frac{4}{2}=2$

Slope of BD$=\frac{-1}{2}$

Eq. of BD

$y+2=\frac{-1}{2}(x+1)$

$2y+4=-x-1$

$\Rightarrow 2y+x=-5$

For co ordinates of D

$\Rightarrow x+2y=-5$

$\Rightarrow 7x+14y=-35$

$AD:7x-y=5$

$15y=-40$

$\Rightarrow y=\frac{-8}{3}$

$x=-5-2y$

$=-5+\frac{16}{3}=\frac{1}{3}$

Co ordinates of $D=(\frac{1}{3},\frac{-8}{3})$