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

Let coordinates of A be $(0, a)$

The diagonals intersect at $P(1, 2)$

We know that the diagonals will be parallel to the

angle bisectors of the two sides $y=x+2$ and $y=7x+3$

i.e. $\frac{x-y+2}{\sqrt{2}}=\pm \frac{7x-y+3}{5\sqrt{2}}$

$\Rightarrow 5x-5y+10=\pm (7x-y+3)$

$\Rightarrow 2x+4y-7=0$ and $12x-6y+13=0$

$\Rightarrow m_1=-1/2$ and $m_2=2$

(where $m_1$ and $m_2$ are the slopes of the given two lines) Let one diagonal be parallel to $2x + 4y- 7 = 0$and other be parallel to $12x-6y+13=0$

The vertex A could be on any of the two diagonals.

Hence, slope of AP is either$-1/2$ or 2.

$\Rightarrow \frac{2-a}{1-0}=2$ or$\frac{2-a}{1-0}=-\frac{1}{2}\Rightarrow a=0$ or$a=\frac{5}{2}$

But $a\ne 0$       $\therefore$                                                              Thus, ordinate of A is 5/2.