# How do you write the standard form of a line given slope= 4/5, passes through (10, -3)?

Jun 20, 2016

$y = \frac{4}{5} x - 11$

#### Explanation:

The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where m represents the slope and b , the y-intercept.

here $m = \frac{4}{5}$

Partial equation is : $y = \frac{4}{5} x + b$

To find b , use the point (10 ,-3) that the line passes through.

x = 10 , y = -3 and substitute into Partial equation.

$\frac{4}{\cancel{5}} ^ 1 \times {\cancel{10}}^{2} + b = - 3 \Rightarrow b = - 3 - 8 = - 11$

$\Rightarrow \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = \frac{4}{5} x - 11} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ is the equation}$

If we multiply both sides by 5 to eliminate the fraction.

$5 y = \frac{4}{\cancel{5}} ^ 1 \times {\cancel{5}}^{1} x - \left(11 \times 5\right)$

$\Rightarrow 5 y = x - 55$

and $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{5 y - x + 55 = 0} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ is another form of the equation}$