The sides of a rhombus $A B C D$ are parallel to the lines, $x-y+2=0$ and $7 x-y+3=0$. If the diagonals of…
The sides of a rhombus $A B C D$ are parallel to the lines, $x-y+2=0$ and $7 x-y+3=0$. 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
2
$\frac{7}{4}$
$\frac{7}{2}$
$\frac{5}{2}$
Solution
Let the coordinate $A$ be $(0, c)$
Equations of the given lines are
$x-y+2=0$ and
$7 x-y+3=0$
We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; $y=x+2$ and $y=7 x+3$
$\therefore$ equation of angle bisectors is given as:
$
\begin{aligned}
&\frac{x-y+2}{\sqrt{2}}=\pm \frac{7 x-y+3}{5 \sqrt{2}} \\
&5 x-5 y+10=\pm(7 x-y+3)
\end{aligned}
$
$\therefore$ Parallel equations of the diagonals are $2 x$ $+4 y-7=0$ and $12 x-6 y+13=0$
$\therefore$ slopes of diagonals are $\frac{-1}{2}$ and 2 .
Now, slope of the diagonal from $A(0, \mathrm{c})$ and passing through $P(1,2)$ is $(2-c)$
$
\therefore 2-c=2 \Rightarrow c=0 \text { (not possible) }
$
$
\therefore 2-c=\frac{-1}{2} \Rightarrow c=\frac{5}{2}
$
$
\therefore \text { ordinate of A is } \frac{5}{2} \text {. }
$