JEE Main 2019 — Straight Lines Question with Solution

Given, the point $P\left(2, 3\right)$ on the line $x+y=7.$

Let, the point of intersection of the two lines be $Q$

Then, $Q$ can be taken as $Q\equiv (\alpha , 7-\alpha )$

Given, $PQ=4$ units and the distance between the points $\left(x_1, y_1\right) \& \left(x_2, y_2\right)$ is $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

$\Rightarrow \sqrt{(\alpha -{2)}^2+{\left(7-\alpha -3\right)}^2}=4$

$\Rightarrow \sqrt{(\alpha -{2)}^2+{\left(4-\alpha \right)}^2}=4$

$\Rightarrow (\alpha -{2)}^2+{\left(4-\alpha \right)}^2=4^2$

$\Rightarrow {\alpha }^2-4\alpha +4+16-8\alpha +{\alpha }^2=16$

$\Rightarrow 2{\alpha }^2-12\alpha +4=0$

$\Rightarrow {\alpha }^2-6\alpha +2=0$

$\Rightarrow \alpha =\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\times 1\times 2}}{2\times 2}$

$\Rightarrow \alpha =\frac{6\pm \sqrt{36-8}}{4}$

$\Rightarrow \alpha =\frac{6\pm 2\sqrt{7}}{4}$

$\Rightarrow \alpha =3\pm \sqrt{7}$

The slope of a line joining the points $\left(x_1, y_1\right) \& \left(x_2, y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}.$

If we take $\alpha =3-\sqrt{7},$ then $Q\equiv \left(3-\sqrt{7}, 4+\sqrt{7}\right)$

And, the slope of $PQ=m=\frac{4+\sqrt{7}-3}{3-\sqrt{7}-2}=\frac{1+\sqrt{7}}{1-\sqrt{7}}$

If we take $\alpha =3+\sqrt{7},$ then $Q\equiv \left(3+\sqrt{7}, 4-\sqrt{7}\right)$

And, the slope of $PQ=m=\frac{4-\sqrt{7}-3}{3+\sqrt{7}-2}=\frac{1-\sqrt{7}}{1+\sqrt{7}}.$