Question 173 of 303MediumMCQ
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:
Solutions · 1
Official solution
We know parametric form of straight line is
\[{{x - {x_1}} \over {\cos \theta }} = {{y - {y_1}} \over {\sin \theta }} = r\]
\[ \Rightarrow \] \[{{x - 2} \over {\cos \theta }} = {{y - 3} \over {\sin \theta }} = 4\]
\[ \therefore \] x = 4 + 2cos\[\theta \]
y = 4 + 3sin\[\theta \]
\[ \therefore \] Point A = (4 + 2cos\[\theta \], 4 + 3sin\[\theta \])
Point A lies on line x + y = 7,
\[ \therefore \] (4 + 2cos\[\theta \]) + (4 + 3sin\[\theta \]) = 7
\[ \Rightarrow \] sin\[\theta \] + cos\[\theta \] = \[{1 \over 2}\]
\[ \Rightarrow \] \[{\sin ^2}\theta + {\cos ^2}\theta \] + 2sin\[\theta \]cos\[\theta \] = \[{1 \over 4}\]
\[ \Rightarrow \] 1 + sin 2\[\theta \] = \[{1 \over 4}\]
\[ \Rightarrow \] sin 2\[\theta \] = \[ - {3 \over 4}\]
\[ \Rightarrow \] \[{{2\tan \theta } \over {1 + {{\tan }^2}\theta }}\] = \[ - {3 \over 4}\]
\[ \Rightarrow \] 3tan2\[\theta \] + 8tan\[\theta \] + 3 = 0
\[ \Rightarrow \] tan\[\theta \] = \[{{ - 8 \pm 2\sqrt 7 } \over 6}\]
So slope = \[{{ - 8 \pm 2\sqrt 7 } \over 6}\]
By checking each options,
Slope = \[{{1 - \sqrt 7 } \over {1 + \sqrt 7 }}\]
As \[{{1 - \sqrt 7 } \over {1 + \sqrt 7 }}\] = \[{{{{\left( {1 - \sqrt 7 } \right)}^2}} \over {1 - 7}}\] = \[{{8 - 2\sqrt 7 } \over { - 6}}\] = \[{{ - 8 + 2\sqrt 7 } \over 6}\]
Answer the question to unlock