JEE Main 2018 — Straight Lines Question with Solution

Let the coordinate $A$ be $(0,c)$ Equations of the given lines are $x-y+2=0$ and $7x-y+3=0$ 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: $$\begin{array}{rl}& \frac{x-y+2}{\sqrt{2}}=\pm \frac{7x-y+3}{5\sqrt{2}}\\ & 5x-5y+10=\pm (7x-y+3)\end{array}$$ $\therefore$ Parallel equations 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 $A(0,\mathrm{c})$ and passing through $P(1,2)$ is $(2-c)$ $$\therefore 2-c=2\Rightarrow c=0\text{ (not possible) }$$ $$\therefore 2-c=\frac{-1}{2}\Rightarrow c=\frac{5}{2}$$ $$\therefore \text{ ordinate of A is }\frac{5}{2}\text{. }$$