Question

The sides of a rhombus $ABCD$ are parallel to the lines, $x-y+2=0$ and $7x-y+3=0$. If the diagonals of the rhombus intersect at $P(1,2)$ and the vertex $A$ (different from the origin) is on the y-axis, then the ordinate of $A$ is

• A. $2$
• B. $\frac{7}{4}$
• C. $\frac{7}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is D $\frac{5}{2}$

Let the coordinate A be $(0,c)$

Equations of the parallel lines are given: $x-y+2=0$ and $7x-y+3=0$

We know that the diagonals will be parallel to the angle bisectors of the two sides $y=x+2$ and $y=7x+3$

$\frac{x-y+2}{\sqrt{2}}=\pm \frac{7x-y+3}{5\sqrt{2}}$

$5x-5y+10=\pm (7x-y+3)$

Parallel equations of the diagonals are $2x+4y-7=0$ and $12x-6y+13=0$

slopes of diagonal $m=\frac{-1}{2}$ and $2$

We know that the slope of diagonal from $A(0,c)$ and passing through$P(1,2)$ is $(2-c)$

therefore $2-c=2⟹c=0$, but it is given that $A$ is not origin, so

$2-c=\frac{-1}{2}⟹c=\frac{5}{2}$

$\therefore$ coordinate of A is $(0,5/2)$