Exam

JEE Main

Year

2025

Date

28 January

Shift

Evening Shift

Subject

Mathematics

Chapter

Straight Lines

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Straight Lines

Two equal sides of an isosceles triangle are along -x+2 y=4 and...

Two equal sides of an isosceles triangle are along $-x+2 y=4$ and $x+y=4$. If m is the slope of its third side, then the sum, of all possible distinct values of $m$, is :

Step-by-step solution

Slope of the third side = slope of the perpendicular bisector of given lines $h: \frac{-x+2 y-4}{\sqrt{5}}= \pm \frac{x+y-4}{\sqrt{2}}$$\begin{aligned} & h_1: \sqrt{2}(-x+2 y-4)=\sqrt{5}(x+y-4) \\ & h_2: \sqrt{2}(-x+2 y-4)=-\sqrt{5}(x+y-4) \\ & M_{L_1}:-\left[\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-2 \sqrt{2}}\right] \\ & M_{L_2}:-\left[\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+2 \sqrt{2}}\right] \\ & M_{L_1}+M_{L_2}=-\left[\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-2 \sqrt{2}}+\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+2 \sqrt{2}}\right]\end{aligned}$$\begin{aligned} & =-\left[\frac{(\sqrt{5}+\sqrt{2})(\sqrt{5}+2 \sqrt{2})+(\sqrt{5}-\sqrt{2})(\sqrt{5}-2 \sqrt{2})}{-3}\right] \\ & =6\end{aligned}$$$