The sides of a rhombus A B C D are parallel to the lines, x-y+2=0...

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