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The length of perpendicular from the point \[(a\cos \alpha ,a\sin \alpha )\]upon the straight line \[y = x\tan \alpha + C,c > 0\]is.
A. \[a\cos \alpha \]
B. \[c{\sin ^2}x\]
C. \[c{\sec ^2}x\]
D. \[c{\cos ^2}x\]

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Last updated date: 25th Jun 2024
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Answer
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Hint: Perpendicular distance say (d) from a given point \[P({x_1},{y_1})\]to a line \[Ax + By + c = 0\] is given as:
\[ \Rightarrow \dfrac{{|A{x_1} + B{y_1} + C|}}{{\sqrt {{A^2} + {B^2}} }}\]. Use this formula to get the answer.

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Complete step-by-step answer:
The given point is P\[(a\cos \alpha ,a\sin \alpha )\].
And our equation of line is\[y = x\tan \alpha + C\].
Perpendicular distance say (d) from a given point\[P({x_1},{y_1})\]to a line \[Ax + By + c = 0\] is given as:
\[ \Rightarrow \dfrac{{|A{x_1} + B{y_1} + C|}}{{\sqrt {{A^2} + {B^2}} }}\].
\[
   \Rightarrow d = \dfrac{{|\left( { - \tan \alpha } \right)(a\cos \alpha ) + (1)(a\sin \alpha ) + c|}}{{\sqrt {{{\left( { - \tan \alpha } \right)}^2} + {{(1)}^2}} }} \\
   \Rightarrow d = \dfrac{{|\left( { - \dfrac{{\sin \alpha }}{{\cos \alpha }}} \right)(a\cos \alpha ) + (a\sin \alpha ) + c|}}{{\sqrt {1 + {{\tan }^2}\alpha } }} \\
   \Rightarrow d = \dfrac{{|( - a\sin \alpha ) + (a\sin \alpha ) + c|}}{{\sqrt {{{\sec }^2}\alpha } }} \\
   \Rightarrow d = \dfrac{c}{{\sec \alpha }} \\
   \Rightarrow d = c\cos \alpha \\
\]
None of the above options is correct.
The required perpendicular distance=\[c\cos \alpha \].

Note: Working formula to solve such questions:
First simplify the equation of straight line.
Then put the points in our given straight line.
Then put the values in \[\dfrac{{|A{x_1} + B{y_1} + C|}}{{\sqrt {{A^2} + {B^2}} }}\]
Where, \[Ax + By + C = 0\] is the equation of straight line and \[({x_1},{y_1})\] are the points from which perpendicular distance is to be found.