Question

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

• A. $(-3,-8)$
• B. $(1/3,-8/3)$
• C. $(-10/3,-7/3)$
• D. $(-3,-9)$

Answer 1

Answer: (B)

$(1/3,-8/3)$

Point of intersection of the two sides is $(1,2)$

$\therefore$ Diagonal passing through $(1,2)$ and $(-1,-2)$ is

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

$\therefore \frac{y-2}{4}=\frac{x-1}{2}$

$\therefore y=2x$ is one diagonal

As diagonals of Rhombus are $\perp$ to each other and also passes through $(-1,-2)$

Its equation is $y=\frac{-1}{2}x+c$

Putting $(-1,-2)$ in equation $c=\frac{-5}{2}$

$\therefore$ equation of other diagonal is $2y+x+5=0$

Solving this equation with $zx-y-5=0$

we get the vertex point $(\frac{1}{3},\frac{-8}{3})$.