﻿ Point-Slope Form
POINT-SLOPE FORM

by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
• PRACTICE (online exercises and printable worksheets)
• Want some other practice with lines?

Suppose a line has slope $\,m\,$ and passes through a known point $\,(x_1,y_1)\,$.
That is, we know the slope of the line and we know a point on the line.

We can get an equation that is ideally suited to these two pieces of information.
This equation is appropriately called the point-slope form of a line.

Here's what to do:

• Recall that $\,(x_1,y_1)\,$ is a known point on a line with slope $\,m\,$.
• Let $\,(x,y)\,$ denote any other point on the line.
• Now, we have two points:   the known point $\,(x_1,y_1)\,$ and a ‘generic’ point $\,(x,y)\,$.
• The slope of the line, computed using these two points, must equal $\,m\,$.
• Using the slope formula, we have:
$\displaystyle\,m = \frac{y-y_1}{x-x_1}\,$
or, equivalently,
$\,y - y_1 = m(x - x_1)\,$
This gives us an extremely useful equation of a line, as summarized below:

POINT-SLOPE FORM line with slope $\,m\,$, passing through $\,(x_1,y_1)\,$
The graph of the equation $$\cssId{s22}{y - y_1 = m(x - x_1)}$$ is a line with slope $\,m\,$ that passes through the point $\,(x_1,y_1)\,$.
Since this equation is ideally suited to the situation where you know a point and a slope,
it is appropriately called point-slope form.
IMPORTANT THINGS TO KNOW ABOUT POINT-SLOPE FORM:
• The variables in the equation $\,y - y_1 = m(x - x_1)\,$ are $\,x\,$ and $\,y\,$.
That is, this is an equation in two variables, $\,x\,$ and $\,y\,$.
Thus, its solution set is the set of all ordered pairs $\,(x,y)\,$ that make it true.
• For a given equation:
• the number $\,m\,$ is a constant (a specific number) that represents the slope of the line;
• the number $\,x_1\,$ (read as ‘ex sub one’) is a constant that represents the $\,x$-value of the known point;
• the number $\,y_1\,$ (read as ‘wye sub one’) is a constant that represents the $\,y$-value of the known point
• As we vary the values of $\,m\,$, $\,x_1\,$, and $\,y_1\,$, we get lots of different equations.
Here are some of them:
 $y - 2 = 5(x - 3)$ ($\,m = 5\,$, $\,x_1 = 3\,$, and $\,y_1 = 2\,$) $y - \frac12 = \sqrt{2}(x - 3.4)$ ($\,m = \sqrt2\,$, $\,x_1 = 3.4\,$, and $\,y_1 = \frac12\,$) $y = 5(x + 1)$ Rewrite the equation as:   $\,y - 0 = 5(x - (-1))\,$ Thus, we see that:   $\,m = 5\,$, $\,x_1 = -1\,$, and $\,y_1 = 0\,$
• So, even though the equation $\,y - y_1 = m(x - x_1)\,$
uses five different ‘letters’
($\,y\,$, $\,y_1\,$, $\,m\,$, $\,x\,$, and $\,x_1\,$) ,
they play very different roles:
• $\,x\,$ and $\,y\,$ are the variables; they determine the nature of the solution set
• $\,m\,$, $\,x_1\,$ and $\,y_1\,$ are called parameters;
they are constant in any particular equation, but vary from equation to equation.
• This is another beautiful example of the power/compactness of the mathematical language!
The single equation $\,y - y_1 = m(x - x_1)\,$ actually describes an entire family of equations,
which has infinitely-many members.
We get the members of this family by choosing real numbers $\,m\,$, $\,x_1\,$ and $\,y_1\,$ to plug in.
• If you know the slope of a line and the $\,y$-intercept, then it's probably easiest to use slope-intercept form.
But, if you know the slope of a line and a point that isn't the $\,y$-intercept, then it's easiest to use point-slope form.
• Remember—just as expressions have lots of different names, so do sentences.
Every non-vertical line can be written in any of these forms:
• point-slope form:   $y - y_1 = m(x - x_1)$
• slope-intercept form:   $y = mx + b$
• general form:   $ax + by + c = 0$
• Here's an example. (Make sure you convince yourself that these are equivalent equations!)
• point-slope form:   $y - 2 = 5(x - 3)$
• slope-intercept form:   $y = 5x -13$
• general form:   $5x - y - 13 = 0$
EXAMPLE:
Question:
Write the point-slope equation of the line with slope $\,5\,$ that passes through the point $\,(3,-2)\,$.
Then, write the line in $\,y = mx + b\,$ form.
Solution:
Here, $\,(x_1,y_1)\,$ is $\,(3,-2)\,$ and $\,m = 5\,$.
Substitution into $\,y - y_1 = m(x-x_1)\,$ gives: $$\cssId{s83}{y - (-2) = 5(x - 3)}$$
 $y$ minus known$y$-value equals knownslope $($ $x$ minus known$x$-value $)$ $y$ $-$ $(-2)$ $=$ $5$ $($ $x$ $-$ $3$ $)$ $y$ $-$ $y_1$ $=$ $m$ $($ $x$ $-$ $x_1$ $)$

Then, put it in slope-intercept form by solving for $\,y\,$:
 $\,y - (-2) = 5(x - 3)\,$ (start with point-slope form) $\,y +2 = 5x - 15\,$ (simplify each side) $\,y = 5x - 17\,$ (subtract $\,2\,$ from both sides)
Master the ideas from this section