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JEE Main 20207 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2020 · 7 January · Evening Shift · Mathematics · Straight Lines

The locus of the mid-points of the perpendiculars drawn from points...

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

Options

  1. A2x−3y=02x - 3y = 0
  2. B5x−7y=05x - 7y = 0
  3. C3x−2y=03x - 2y = 0
  4. D7x−5y=07x - 5y = 0

Correct answer

B. 5x−7y=05x - 7y = 0

Step-by-step solution

The locus of the mid-points of the perpendiculars drawn from points on the line x = 2y , to the line x = y , is.
Slope of PQ=k−αh−2α=−1PQ = \dfrac{k - \alpha}{h - 2\alpha} = - 1
k−α=−h+2α⇒α=h+k3…(1)<br/>Also,k - \alpha = - h + 2\alpha ⇒ \alpha = \dfrac{h + k}{3} \ldots \left(1\right)<br />Also,2h = 2\alpha + \betaandand2k = \alpha + \beta<br/><br />2h = \alpha + 2k⇒ \alpha = 2h - 2k \ldots \left(2\right)<br/>From(1) and  <br />From \left(1\right) \text{\text{ and } }\left(2\right),
\dfrac{h + k}{3} =2(h−k)2\left(h - k\right)
So, locus is 6x−6y=x+y⇒5x=7y6x - 6y = x + y ⇒ 5x = 7y.
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