# How do you write the equation of a line in point slope form and slope intercept form given point (6, -3) and has a slope of 1/2?

Jun 1, 2018

Point-Slope Form is
$y + 3 = \frac{1}{2} \left(x - 6\right)$

Slope-Intercept Form is
$y = \frac{1}{2} x - 6$

#### Explanation:

The point-slope form of the equation of a line is

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

Where $m$ is the slope and the point is $\left({x}_{1} , {y}_{1}\right)$

For this problem

$m = \frac{1}{2}$
${x}_{1} = 6$
${y}_{1} = - 3$

Plug in the values

$y - \left(- 3\right) = \frac{1}{2} \left(x - 6\right)$

Simplify the signs

$y + 3 = \frac{1}{2} \left(x - 6\right)$
This is the point-slope form of the equation

Now solve for $y$ to get the slope-intercept form.

$y + 3 = \frac{1}{2} \left(x - 6\right)$

Use the distributive property to eliminate the parenthesis

$y + 3 = \frac{1}{2} x - 3$

Now isolate the $y$ using the additive inverse

$y \cancel{+ 3} \cancel{- 3} = \frac{1}{2} x - 3 - 3$

$y = \frac{1}{2} x - 6$
This is the slope-intercept form of the equation

Jun 1, 2018

See a solution process below:

#### Explanation:

The point-slope form of a linear equation is:

$\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$

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

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

$\left(y - \textcolor{b l u e}{- 3}\right) = \textcolor{red}{\frac{1}{2}} \left(x - \textcolor{b l u e}{6}\right)$

$\left(y + \textcolor{b l u e}{3}\right) = \textcolor{red}{\frac{1}{2}} \left(x - \textcolor{b l u e}{6}\right)$

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.

We can solve the point-slope equation for $y$ giving:

$\left(y + \textcolor{b l u e}{3}\right) = \textcolor{red}{\frac{1}{2}} \left(x - \textcolor{b l u e}{6}\right)$

$y + \textcolor{b l u e}{3} = \left(\textcolor{red}{\frac{1}{2}} \times x\right) - \left(\textcolor{red}{\frac{1}{2}} \times \textcolor{b l u e}{6}\right)$

$y + \textcolor{b l u e}{3} = \frac{1}{2} x - \frac{6}{2}$

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

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

$y + 0 = \frac{1}{2} x - 6$

$y = \textcolor{red}{\frac{1}{2}} x - \textcolor{b l u e}{6}$