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

Answer 1

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

Let the point $A$ be $(0,a)$
As we know, diagonals of rhombus will be parallel to the bisector of the lines which are parallel to sides of rhombus
So, bisector of the lines $x-y+2=0$ and $7x-y+3=0$ are
$\frac{x-y+2}{\sqrt{2}}=\pm \frac{7x-y+3}{5\sqrt{2}}$
$\Rightarrow 5x-5y+10=\pm (7x-y+3)$
So, bisectors are $2x+4y-7=0\text{and}12x-6y+13=0$
$\therefore$ Slope of diagonals $=-\frac{1}{2}\text{and}2$
As, slope of $AP=2-a$
$\Rightarrow 2-a=-\frac{1}{2}\text{or}2-a=2$
$\Rightarrow a=\frac{5}{2}\text{or}a=0$
$a=0$ can not be possible because vertex $A$ should be different from the origin.
Hence, $a=\frac{5}{2}$