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Find the Equation of a Line Passing Through (3, −2) and Perpendicular to the Line X − 3y + 5 = 0.

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प्रश्न

Find the equation of a line passing through (3, −2) and perpendicular to the line x − 3y + 5 = 0.

संक्षेप में उत्तर
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उत्तर

The equation of the line perpendicular to x − 3y + 5 = 0 is \[3x + y + \lambda = 0\], where

\[\lambda\] is a constant.

It passes through (3, −2).

\[9 - 2 + \lambda = 0\]

\[ \Rightarrow \lambda = - 7\]

Substituting  \[\lambda\] = −7 in \[3x + y + \lambda = 0\], we get 

\[3x + y - 7 = 0\] , which is the required line.

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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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अध्याय 23: The straight lines - Exercise 23.12 [पृष्ठ ९२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 23 The straight lines
Exercise 23.12 | Q 2 | पृष्ठ ९२

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