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QJEEM19Ma7c41: JEE Main 2019 Mathematics - Class 11 JEEM19Ma7c41 Solution: Slope of a line passing through P(2, 3) and intersecting the…

Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is :

Multiple Choice Options

  • A 7−17+1{{\sqrt 7 - 1} \over {\sqrt 7 + 1}}
  • B 5−15+1{{\sqrt 5 - 1} \over {\sqrt 5 + 1}}
  • C 1−51+5{{1 - \sqrt 5 } \over {1 + \sqrt 5 }}
  • D 1−71+7{{1 - \sqrt 7 } \over {1 + \sqrt 7 }}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 100 English Explanation

We know parametric form of straight line is

x−x1cos⁡θ=y−y1sin⁡θ=r{{x - {x_1}} \over {\cos \theta }} = {{y - {y_1}} \over {\sin \theta }} = r

⇒\Rightarrow x−2cos⁡θ=y−3sin⁡θ=4{{x - 2} \over {\cos \theta }} = {{y - 3} \over {\sin \theta }} = 4

∴\therefore x = 4 + 2cos θ\theta

y = 4 + 3sin θ\theta

∴\therefore Point A = (4 + 2cos θ\theta , 4 + 3sin θ\theta )

Point A lies on line x + y = 7,

∴\therefore (4 + 2cos θ\theta ) + (4 + 3sin θ\theta ) = 7

⇒\Rightarrow sin θ\theta + cos θ\theta = 12{1 \over 2}

⇒\Rightarrow sin⁡2θ+cos⁡2θ{\sin ^2}\theta + {\cos ^2}\theta + 2sin θ\theta cos θ\theta = 14{1 \over 4}

⇒\Rightarrow 1 + sin 2 θ\theta = 14{1 \over 4}

⇒\Rightarrow sin 2 θ\theta = −34- {3 \over 4}

⇒\Rightarrow 2tan⁡θ1+tan⁡2θ{{2\tan \theta } \over {1 + {{\tan }^2}\theta }} = −34- {3 \over 4}

⇒\Rightarrow 3tan2 θ\theta + 8tan θ\theta + 3 = 0

⇒\Rightarrow tan θ\theta = −8±276{{ - 8 \pm 2\sqrt 7 } \over 6}

So slope = −8±276{{ - 8 \pm 2\sqrt 7 } \over 6}

By checking each options,

Slope = 1−71+7{{1 - \sqrt 7 } \over {1 + \sqrt 7 }}

As 1−71+7{{1 - \sqrt 7 } \over {1 + \sqrt 7 }} = (1−7)21−7{{{{\left( {1 - \sqrt 7 } \right)}^2}} \over {1 - 7}} = 8−27−6{{8 - 2\sqrt 7 } \over { - 6}} = −8+276{{ - 8 + 2\sqrt 7 } \over 6}

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