The sides of a rhombs ABCD is parallel to the lines x – y + 2 = 0 and 7x – y + 3 = 0. If the diagonals of the rhombus intersects at P (1, 2) and the vertex A (different from origin ) is on the y-axis. Then the ordinate of A is _____

Let  $A\left(0,\alpha \right)$be the point and P(1,  2 )

The sides $\overline{AB}and\overline{AD}$ are parallel to the lines $x-y+2=0and7x-y+3=0$

The diagonal AC is parallel to the bisector of angle b/w these two lines

Equation of bisectors are

  $\begin{array}{l}\frac{x-y+2}{\sqrt{2}}=\pm \frac{7x-y+3}{\sqrt{50}}\\ \Rightarrow 5\left(x-y+2\right)=\pm \left(7x-y+3\right)\\ \Rightarrow 5x-5y+10=7x-y+3,5x-5y+10=-7x+y-3\\ \Rightarrow 2x+4y-7=0,12x-6y+13=0\end{array}$

Slope of AP = -1/2  or  2

$\begin{array}{l}Slopeof\overline{AP}=\frac{2-\alpha }{1-0}=2-\alpha \\ If2-\alpha =\frac{-1}{2}, If2-\alpha =2\\ \alpha =2+\frac{1}{2}, \alpha =2-2\\ \alpha =\frac{5}{2}, \alpha \ne 0\left(Givendiffrentfromorigin\right)\\ \therefore A\left(0,\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\\ \alpha =\frac{5}{2}=2.5\end{array}$