# How do you write an equation of a line passing through (10, 5), perpendicular to 5x+4y=8?

Apr 11, 2017

See the entire solution process.

#### Explanation:

The line given in the problem is in Standard Form. The standard form of a linear equation is: $\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$

Where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

$\textcolor{red}{5} x + \textcolor{b l u e}{4} y = \textcolor{g r e e n}{8}$

The slope of an equation in standard form is: $m = - \frac{\textcolor{red}{A}}{\textcolor{b l u e}{B}}$. Substituting the values from the equation in the problem gives:

$m = - \frac{\textcolor{red}{5}}{\textcolor{b l u e}{4}}$

Let's call the slope of the line perpendicular to this line ${m}_{p}$. The slope of a perpendicular line is:

${m}_{p} = - \frac{1}{m} = \frac{4}{5}$

We can now use the point-slope formula to find the equation of the line. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$

Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.

Substituting the slope we calculated and the values from the point in the problem gives:

$\left(y - \textcolor{red}{5}\right) = \textcolor{b l u e}{\frac{4}{5}} \left(x - \textcolor{red}{10}\right)$

We can convert this to the slope-intercept form by solving for $y$. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y - \textcolor{red}{5} = \left(\textcolor{b l u e}{\frac{4}{5}} \times x\right) - \left(\textcolor{b l u e}{\frac{4}{5}} \times \textcolor{red}{10}\right)$

$y - \textcolor{red}{5} = \frac{4}{5} x - \frac{40}{5}$

$y - \textcolor{red}{5} = \frac{4}{5} x - 8$

$y - \textcolor{red}{5} + 5 = \frac{4}{5} x - 8 + 5$

$y - 0 = \frac{4}{5} x - 3$

$y = \textcolor{red}{\frac{4}{5}} x - \textcolor{b l u e}{3}$