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 pointslope 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:
$\displaystyle\,m = \frac{yy_1}{xx_1}\,$
or, equivalently,
$\,y  y_1 = m(x  x_1)\,$
This gives us an extremely useful equation of a line, as summarized below:
POINTSLOPE 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 pointslope form.
IMPORTANT THINGS TO KNOW ABOUT POINTSLOPE 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 infinitelymany 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 slopeintercept 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 pointslope form.

Remember—just as expressions have lots of different names, so do sentences.
Every nonvertical line can be written in any of these forms:

pointslope form:
$y  y_1 = m(x  x_1)$

slopeintercept form:
$y = mx + b$

general form:
$ax + by + c = 0$

Here's an example.
(Make sure you convince yourself that these are equivalent equations!)

pointslope form:
$y  2 = 5(x  3)$

slopeintercept form:
$y = 5x 13$

general form:
$5x  y  13 = 0$
EXAMPLE:
Question:
Write the pointslope 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 $\,y  y_1 = m(xx_1)\,$ gives:
$$
\cssId{s83}{y  (2) = 5(x  3)}
$$
$y$ 
minus 
known $y$value 
equals 
known slope 
$($ 
$x$ 
minus 
known $x$value 
$)$ 
$y$ 
$$ 
$(2)$ 
$=$ 
$5$ 
$($ 
$x$ 
$$ 
$3$ 
$)$ 
$y$ 
$$ 
$y_1$ 
$=$ 
$m$ 
$($ 
$x$ 
$$ 
$x_1$ 
$)$ 
Then, put it in slopeintercept form by solving for $\,y\,$:
$\,y  (2) = 5(x  3)\,$

(start with pointslope form) 
$\,y +2 = 5x  15\,$ 
(simplify each side) 
$\,y = 5x  17\,$ 
(subtract $\,2\,$ from both sides) 
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Horizontal and Vertical Lines