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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
$2$
$\frac{7}{4}$
$\frac{7}{2}$
$\frac{5}{2}$
Solution
Let the coordinate $A$ be $(0,c)$
Equations of the given lines are
$x-y+2=0$
$7x-y+3=0$
we know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; $y=x+2$ and $y=7x+3$
$\therefore $ equation of angle bisectors is given as:
$\frac{{x – y + 2}}{{\sqrt 2 }} = \pm \frac{{7x – y + 3}}{{5\sqrt 2 }}$
$5x – 5y + 10 = \pm \left( {7x – y + 3} \right)$
$\therefore $Parallel equation of the diagonals are $2x+4y-7=0$ and $12x-6y+13=0$
$\therefore $ slopes of diagonals are $\frac{{ – 1}}{2}$ and $2$.
Now, slope of the diagonal from $A(0,c)$ and passing through $P(1,2)$ is $(2-c)$
$\therefore 2 – c = 2 \Rightarrow c = 0$ (not possible)
$\therefore 2 – c = \frac{{ – 1}}{2} \Rightarrow c = \frac{5}{2}$
