Question

Two sides of a rhombus $ABCD$ are parallel to the lines $y=x+2$ and $y=7x+3$. If the diagonals of the rhombus intersect at the point $(1,2)$ and the vertex $A$ is on the Y-axis at a distance $a$ unit from the origin, then $a$ is equal to:

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

Answer 1

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

Let the coordinate of $A$ be $(0,\alpha )$.
Since the sides $AB$ and $AD$ are parallel to the lines $y=x+2$ and $y=7x+3$ respectively.
Therefore the diagonal $AC$ is parallel to the bisector of the angle between these two lines.
The equation of the bisectors are given by,

$\frac{x-y+2}{\sqrt{2}}=\pm \frac{7x-y+3}{\sqrt{50}}\Rightarrow 5(x-y+2)=\pm (7x-y+3)$
$\Rightarrow 2x+4y-7=0$ and $12x-6y+13=0$
Thus, the diagonals of the rhombus are parallel to the lines
$2x+4y-7=0$ and $12x-6y+14=0$
Therefore slope of $AE=-\frac{2}{4}$ or $\frac{12}{6}$
$\Rightarrow \frac{2-\alpha }{1-0}=-\frac{1}{2}$ or $\frac{2-\alpha }{1-0}=2$
$\Rightarrow \alpha =\frac{5}{2}$ or $\alpha =0$