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JEE Main 202118 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2021 · 18 March · Morning Shift · Mathematics · Straight Lines

The equation of one of the straight lines which passes through the...

The equation of one of the straight lines which passes through the point (1,3)( 1, 3 ) and makes an angles tan⁡−1(2)\tan^{- 1} ( \sqrt{2} ) with the straight line, y+1=32xy + 1 = 3\sqrt{2}x is

Options

  1. A42x+5y−(15+42)=04\sqrt{2}x + 5y - ( 15 + 4\sqrt{2} ) = 0
  2. B52x+4y−(15+42)=05\sqrt{2}x + 4y - ( 15 + 4\sqrt{2} ) = 0
  3. C42x+5y−42=04\sqrt{2}x + 5y - 4\sqrt{2} = 0
  4. D42x−5y−(5+42)=04\sqrt{2}x - 5y - ( 5 + 4\sqrt{2} ) = 0

Correct answer

A. 42x+5y−(15+42)=04\sqrt{2}x + 5y - ( 15 + 4\sqrt{2} ) = 0

Step-by-step solution

y=mx+cy = mx + c

3=m+c3 = m + c

2=∣m−321+32m∣\sqrt{2} = \left|\dfrac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m}\right|

⇒6m+2=±(m−32)⇒ 6m + \sqrt{2} = \pm \left(m - 3\sqrt{2}\right)

from Positive sign

⇒5m=−42⇒m=−425⇒ 5m = - 4\sqrt{2} ⇒ m = \dfrac{- 4\sqrt{2}}{5}

from negative sign

⇒6m+2=−m+32⇒ 6m + \sqrt{2} = - m + 3\sqrt{2}

⇒7m=22⇒m=227⇒ 7m = 2\sqrt{2} ⇒ m = \dfrac{2\sqrt{2}}{7}

According to options take m=−425m = \dfrac{- 4\sqrt{2}}{5}

So y=−42x5+3+425y = \dfrac{- 4\sqrt{2}x}{5} + \dfrac{3 + 4\sqrt{2}}{5}

42x+5y−(15+42)=04\sqrt{2}x + 5y - ( 15 + 4\sqrt{2} ) = 0

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