Point-Slope Form

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 \((6,-3)\):

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