# How do you write an equation of a line passing through (-3, 5), perpendicular to 4x + 2y = 5?

Apr 16, 2017

$\text{the equation of line can be written as : }$
$\textcolor{b l u e}{y = \frac{1}{2} x + \frac{13}{2}} \text{ or }$
$\textcolor{b l u e}{- x + 2 y = 13}$

#### Explanation: $4 x + 2 y = 5$

$\text{First,let us rearrange the equation above to find slope of the line.}$

$\textcolor{red}{2 y = - 4 x + 5}$

$\textcolor{red}{y = - \frac{4 x}{2} + \frac{5}{2}}$

$\textcolor{red}{y = - 2 x + \frac{5}{2}}$

$\text{in equation y=m x+n,the coefficient of the term x gives}$
$\text{the slope of the line.}$

$\text{The slope of the red line is } {m}_{r} = - 2.$

$\text{İf two line are perpendicular each other,the product of their }$$\text{slopes is -1.}$

${m}_{b} \cdot {m}_{r} = - 1$

${m}_{b} : \text{the slope of the blue line}$
${m}_{r} : \text{the slope of the red line.}$

${m}_{b} \cdot \left(- 2\right) = - 1$

${m}_{b} = \frac{1}{2}$

$\text{Since the slope and one point is known,the equation of the blue}$$\text{line can be written.}$

y-y_1=m(x-x_1

$m = {m}_{b} = \frac{1}{2}$

$P \left(- 3 , 5\right) \text{ ; "x_1=-3" , } {y}_{1} = 5$

$\textcolor{b l u e}{y - 5 = \frac{1}{2} \left(x + 3\right)}$

$\textcolor{b l u e}{y = \frac{1}{2} \left(x + 3\right) + 5}$

$\textcolor{b l u e}{y = \frac{1}{2} x + \frac{3}{2} + 5}$

$\textcolor{b l u e}{y = \frac{1}{2} x + \frac{13}{2}}$

$\text{Or }$

$\textcolor{b l u e}{- x + 2 y = 13}$