# Find the Equation of the Line Passing Through the Point (−3, 5) and Perpendicular to the Line Joining (2, 5) and (−3, 6). - Mathematics

Find the equation of the line passing through the point (−3, 5) and perpendicular to the line joining (2, 5) and (−3, 6).

#### Solution

The given points are $A \left( 2, 5 \right) \text { and } B \left( - 3, 6 \right)$.

$\therefore$ Slope of AB $= \frac{6 - 5}{- 3 - 2} = - \frac{1}{5}$

Let m be the slope of the required line. Then,

$m \times \text { Slope of } AB = - 1$

$\Rightarrow m \times \frac{- 1}{5} = - 1$

$\Rightarrow m = 5$

So, the equation of the line that passes through (−3, 5) and has slope 5 is

$y - 5 = 5\left( x + 3 \right)$

$\Rightarrow 5x - y + 20 = 0$

Hence, the equation of the required line is

$5x - y + 20 = 0$

Concept: Straight Lines - Equation of Family of Lines Passing Through the Point of Intersection of Two Lines
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.4 | Q 14 | Page 29