Point-Slope Form

1. Complete the point-slope equation of the line through $(1,3)$ and $(5,1)$.
Use exact numbers.

$y-3= \text { ______ }$

2. Complete the point-slope equation of the line through $(-8, -8)$ and $( -7,9)$.
Use exact numbers.

$y-9= \text { ______ }$

3. Complete the point-slope equation of the line through $(-1, -10)$ and $( 5, 2)$.
Use exact numbers.

$y-2= \text { ______ }$

4. Complete the point-slope equation of the line through $(1, 0)$ and $( 6, -3)$.
Use exact numbers.

$y-(-3)= \text { ______ }$

1. $y-3=-\frac{1}{2}(x-1)$

The general point-slope form $y-y_{1}=m\left(x-x_{1}\right)$, where $m$ is the slope and $\left(x_{1}, y_{1}\right)$ is a point on the line.

Why?

The point-slope form comes directly from the definition of slope. We'll use $m$ to represent slope and $\left(x_{1}, y_{1}\right)$ as one point on the graph of the equation.

$\frac{\text { change in } y}{\text { change in } x}=m$

$\begin{aligned} &\frac{y-y_{1}}{x-x_{1}}=m \\ &y-y_{1}=m\left(x-x_{1}\right) \end{aligned}$

Let's find the slope between $(1, 3)$ and $(5, 1)$:

$\begin{aligned} \text { Slope } &=\frac{1-3}{5-1} \\ &=\frac{-2}{4} \\ &=-\frac{1}{2} \end{aligned}$

The incomplete equation starts with $y-3$, so we need to use the point $(1,3)$:

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

2. $y-9=17(x-(-7))$

The general point-slope form is $y-y_{1}=m\left(x-x_{1}\right)$, where $m$ is the slope and $\left(x_{1}, y_{1}\right)$ is a point on the line.

Why?

The point-slope form comes directly from the definition of slope. We'll use $m$ to represent slope and $\left(x_{1}, y_{1}\right)$ as one point on the graph of the equation.

$\frac{\text { change in } y}{\text { change in } x}=m$

$\begin{aligned} &\frac{y-y_{1}}{x-x_{1}}=m \\ &y-y_{1}=m\left(x-x_{1}\right) \end{aligned}$

Let's find the slope between $(-7,9)$ and $(-8, -8)$:

$\begin{aligned} \text { Slope } &=\frac{9-(-8)}{-7-(-8)} \\ &=\frac{17}{1} \\ &=17 \end{aligned}$

The incomplete equation starts with $y-9$, so we need to use the point $(-7, 9)$:

$y-9=17(x-(-7))$

3. $y-2=2(x-5)$ 

The general point-slope form is $y-y_{1}=m\left(x-x_{1}\right)$, where $m$ is the slope and $\left(x_{1}, y_{1}\right)$ is a point on the line.

Why?

The point-slope form comes directly from the definition of slope. We'll use $m$ to represent slope and $\left(x_{1}, y_{1}\right)$ as one point on the graph of the equation.

$\frac{\text { change in } y}{\text { change in } x}=m$

$\begin{aligned} &\frac{y-y_{1}}{x-x_{1}}=m \\ &y-y_{1}=m\left(x-x_{1}\right) \end{aligned}$

Let's find the slope between $(5,2)$ and $( -1,-10)$:

$\begin{aligned} \text { Slope } &=\frac{2-(-10)}{5-(-1)} \\ &=\frac{12}{6} \\ &=2 \end{aligned}$

The incomplete equation starts with $y-2$, so we need to use the point $(5,2)$:

$y-2=2(x-5)$

4. $y-(-3)=-\frac{3}{5}(x-6)$

The general point-slope form is $y-y_{1}=m\left(x-x_{1}\right)$, where $m$ is the slope and $\left(x_{1}, y_{1}\right)$ is a point on the line.

Why?

The point-slope form comes directly from the definition of slope. We'll use $m$ to represent slope and $\left(x_{1}, y_{1}\right)$ as one point on the graph of the equation.

$\frac{\text { change in } y}{\text { change in } x}=m$

$\begin{aligned} &\frac{y-y_{1}}{x-x_{1}}=m \\ &y-y_{1}=m\left(x-x_{1}\right) \end{aligned}$

Let's find the slope between $(6, -3)$ and $( 1, 0)$:

$\begin{aligned} \text { Slope } &=\frac{-3-0}{6-1} \\ &=\frac{-3}{5} \\ &=-\frac{3}{5} \end{aligned}$

The incomplete equation starts with $y -(-3)$, so we need to use the point $(6,-3)$:

$y-(-3)=-\frac{3}{5}(x-6)$