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, coordinates of $A$ can be

• A. $(0,\frac{3}{2})$
• B. $(0,0)$
• C. $(0,\frac{5}{2})$
• D. $(0,\frac{2}{5})$

Answer 1

Answer: (B), (C)

$(0,0)$
$(0,\frac{5}{2})$

Let the coordinates of $A$ be $(0,\alpha )$.
Since the sides $AB$ ands $AD$ 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$\frac{A_{1}x+B_{1}y+C_1}{\sqrt{A_1^2+B_1^2}}=\pm \frac{A_{2}x+B_{2}y+C_2}{\sqrt{A_2^2+B_2^2}}$

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 $12x-6y+13=0$
$\therefore$ 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 $\alpha =0$