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
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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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