Verified Solution • 2025–26 Session moderate

QJEEM20Ma435b: JEE Main 2020 Mathematics - Class 11 JEEM20Ma435b Solution: The locus of the mid-points of the perpendiculars drawn from…

The locus of the mid-points of the perpendiculars drawn from points on the line, x = 2y to the line x = y is :

Step-by-Step Verified Solution

Slope of line y = x is 1

Line AB is perpendicular to line y = x so

Slope of AB = -1

Also slope of AB =

\frac{α - β}{2 α - β}

∴

\frac{α - β}{2 α - β}

= -1

\Rightarrow

α

= 2

β

h =

\frac{2 α + β}{2}

\Rightarrow

2h =

\frac{4 α + 2 β}{2}

\frac{4 α + 3 α}{2}

\frac{7 α}{2}

Also k =

\frac{α + β}{2}

\Rightarrow

2k =

\frac{5 α}{2}

\frac{h}{k} = \frac{7}{5}

\Rightarrow

5h = 7k

\Rightarrow

5x = 7y