JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 5)

The locus of the mid-points of the perpendiculars drawn from points on the line, x = 2y to the line x = y is :
3x - 2y = 0
7x - 5y = 0
2x - 3y = 0
5x - 7y = 0

Explanation

JEE Main 2020 (Online) 7th January Evening Slot Mathematics - Straight Lines and Pair of Straight Lines Question 94 English Explanation

Slope of line y = x is 1

Line AB is perpendicular to line y = x so

Slope of AB = -1

Also slope of AB = $${{\alpha - \beta } \over {2\alpha - \beta }}$$

$$ \therefore $$ $${{\alpha - \beta } \over {2\alpha - \beta }}$$ = -1

$$ \Rightarrow $$ 3$$\alpha $$ = 2$$\beta $$

h = $${{2\alpha + \beta } \over 2}$$

$$ \Rightarrow $$ 2h = $${{4\alpha + 2\beta } \over 2}$$ = $${{4\alpha + 3\alpha } \over 2}$$ = $${{7\alpha } \over 2}$$

Also k = $${{\alpha + \beta } \over 2}$$

$$ \Rightarrow $$ 2k = $${{5\alpha } \over 2}$$

So $${h \over k} = {7 \over 5}$$

$$ \Rightarrow $$ 5h = 7k

$$ \Rightarrow $$ 5x = 7y

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