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{{ printedBook.courseTrack.name }} {{ printedBook.name }} The equation of a line can be written in slope-intercept form or standard form. However, when the slope and a point are given, point-slope form is another alternative.

$y-y_1 = m(x-x_1)$

As its name suggests, point-slope form requires a point on and the slope of the line.

To write the point-slope form for the line that passes through the points $(\text{-}1,5) \quad \text{and} \quad (1,1),$ find the slope, then use either point to write the equation.Find the slope

$m = \dfrac{y_2-y_1}{x_2-x_1}$

SubstitutePointsSubstitute $\left({\color{#0000FF}{1,1}}\right)$ & $\left({\color{#009600}{\text{-}1,5}}\right)$

$m = \dfrac{{\color{#0000FF}{1}}-{\color{#009600}{5}}}{{\color{#0000FF}{1}}-\left( {\color{#009600}{\text{-}1}} \right)}$

SubTermSubtract term

$m=\dfrac{\text{-}4}{1-(\text{-}1)}$

SubNeg$a-(\text{-} b)=a+b$

$m=\dfrac{\text{-}4}{2}$

CalcQuotCalculate quotient

$m=\text{-}2$

Choose one point on the line

Substitute values

Write the equation for the line in point-slope form.

Show Solution

To write the equation of the line in point-slope form, $y-y_1=m(x-x_1),$ we need the slope and a point.

Since we're given the line as a graph we can use the rise and run to find the slope. We'll find the rise and run between two arbitratily chosen points, the $y$-intercept, $(0,\text{-} 2),$ and $(2,1).$

Here, the run is $2$ and the rise is $3,$ which gives a slope of $m=\dfrac{3}{2}.$

Next, any point on the line can be used for $(x_1,y_1).$ Let's use the same point as above, $(2,1).$

Lastly, substitute the found values of $m$ and $(x_1,y_1)$ into $y-y_1=m(x-x_1).$ Here, $m={\color{#0000FF}{\frac{3}{2}}}$ and $(x_1,y_1)=({\color{#009600}{2}},{\color{#009600}{1}})$ will be substituted. $y-{\color{#009600}{1}}={\color{#0000FF}{\dfrac{3}{2}}}(x-{\color{#009600}{2}}).$

All points in the table lie on the same line.

Write the equation of the line in point-slope form.

Show Solution

To write an equation in the form $y-y_1=m(x-x_1),$ the slope of the line, $m,$ and any point on the line, $(x_1,y_1),$ must be known. We'll begin by finding the slope between consecutive points. It can be seen that the difference between each $x$-value is $1.$

In the right column, we see that the difference between each value is $2.$

Therefore, $m=2.$ Next, we need any point on the line. We know that all points in the table are on the line, so we can choose any of them. Let's use $(\text{-}5,\text{-}1)$ to be $(x_1,y_1.)$ Substituting this point and $m=2$ gives us the line.$y-y_1=m(x-x_1)$

Substitute$m={\color{#0000FF}{2}}$

$y-y_1={\color{#0000FF}{2}}(x-x_1)$

$y-({\color{#009600}{\text{-}1}})=2(x-({\color{#0000FF}{\text{-}5}}))$

SubNeg$a-(\text{-} b)=a+b$

$y+1=2(x+5)$

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