Find the equation through (2,3) and perpendicular to the line 3x+4y-5=0.

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Solution

Given eqn. of line : 3x+4y -5=0
Slope of given line $(= m_{1}) = \frac{- 3}{4}$ $(= \frac{- c o e f f . o f x}{c o e f f o f y})$
Let, $^{'} m_{2}^{'}$ be the slope of line h to given line
Using the ralation, $m_{1} . m_{2} = - 1$
$\Rightarrow \frac{- 3}{4} . m_{2} = - 1 \Rightarrow (m_{2} = \frac{4}{3})$
Also, if pares through (2,3)
$∴$ equation of line = $\frac{y - y_{1}}{x - x_{1}} = m_{2}$
$\Rightarrow \frac{y - 3}{x - 2} = \frac{4}{3} \Rightarrow 3 y - 9 = 4 x - 8 \Rightarrow (3 y - 4 x - 1 = 0)$

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