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The Equation of the Line Passing Through (1, 5) and Perpendicular to the Line 3x − 5y + 7 = 0 is

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Question

The equation of the line passing through (1, 5) and perpendicular to the line 3x − 5y + 7 = 0 is

Options

  •  5x + 3y − 20 = 0

  • 3x − 5y + 7 = 0

  • 3x − 5y + 6 = 0

  • 5x + 3y + 7 = 0

MCQ
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Solution

 5x + 3y − 20 = 0

A line perpendicular to 3x − 5y + 7 = 0 is given by \[5x + 3y + \lambda = 0\]

This line passes through (1, 5).

\[5 + 15 + \lambda = 0\]

\[ \Rightarrow \lambda = - 20\]

Therefore, the equation of the required line is \[5x + 3y - 20 = 0\]

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Equations of Line in Different Forms - Equation of Family of Lines Passing Through the Point of Intersection of Two Lines
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Chapter 23: The straight lines - Exercise 23.21 [Page 134]

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R.D. Sharma Mathematics [English] Class 11
Chapter 23 The straight lines
Exercise 23.21 | Q 19 | Page 134

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