The sides of a rhombus A B C D are parallel to the lines, x-y+2=0 and 7 x-y+3=0. If the diagonals of the rhombus intersect at P(1,2) and the vertex A…

Let the coordinate $A$A be $(0, c)$(0,c) Equations of the given lines are $x-y+2=0$x−y+2=0 and $7 x-y+3=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; $y=x+2$y=x+2 and $y=7 x+3$y=7x+3 $\therefore$∴ equation of angle bisectors is given as: $$\begin{aligned} &\frac{x-y+2}{\sqrt{2}}=\pm \frac{7 x-y+3}{5 \sqrt{2}} \\ &5 x-5 y+10=\pm(7 x-y+3) \end{aligned}$$​2​x−y+2​=±52​7x−y+3​5x−5y+10=±(7x−y+3)​ $\therefore$∴ Parallel equations of the diagonals are $2 x$2x $+4 y-7=0$+4y−7=0 and $12 x-6 y+13=0$12x−6y+13=0 $\therefore$∴ slopes of diagonals are $\frac{-1}{2}$2−1​ and 2 . Now, slope of the diagonal from $A(0, \mathrm{c})$A(0,c) and passing through $P(1,2)$P(1,2) is $(2-c)$(2−c) $$\therefore 2-c=2 \Rightarrow c=0 \text { (not possible) }$$∴2−c=2⇒c=0 (not possible)  $$\therefore 2-c=\frac{-1}{2} \Rightarrow c=\frac{5}{2}$$∴2−c=2−1​⇒c=25​ $$\therefore \text { ordinate of A is } \frac{5}{2} \text {. }$$∴ ordinate of A is 25​.