FINDING EQUATIONS OF LINES
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!

Every non-vertical line in the coordinate plane can be described by an equation of the form $\,y = mx + b\,$, where:

The equation $\,y = mx + b\,$ is called the slope-intercept form of the line.

Two different points uniquely determine a line.
One point and a slope also uniquely determine a line.
This web exercise gives you practice writing the equation of the line in these two situations.

EXAMPLE (KNOWN POINT, KNOWN SLOPE)
Question:
Find the equation of the line with slope $\,3\,$ that passes through the point $\,(-1,5)\,$.
Write the equation in $\,y = mx + b\,$ form.
Solution:
 $y = mx + b$ (A line with slope $\,3\,$ isn't vertical, so it can be described by an equation of this form.) $y = 3x + b$ (Substitute the known slope, $\,3\,$, in for $\,m\,$. Next, we must find $\,b\,$.) $5 = 3(-1) + b$ (Since $\,(-1,5)\,$ lies on the line, substitution of $\,-1\,$ for $\,x\,$ and $\,5\,$ for $\,y\,$ makes the equation true.) $5 = -3 + b$ (simplify) $\,b = 8\,$ (add $\,3\,$ to both sides; write in the conventional way) $y = 3x + 8$ (substitute the now-known value of $\,b\,$ into the equation)
Thus, the line with slope $\,3\,$ that passes through $\,(-1,5)\,$ is described by the equation $\,y = 3x + 8\,$.

Make sure you understand what this means!
Let $\,\ell\,$ denote the line with slope $\,3\,$ that passes through the point $\,(-1,5)\,$.
Every point that lies on $\,\ell\,$ has coordinates that make the equation $\,y = 3x + 8\,$ true.
Every point that doesn't lie on $\,\ell\,$ has coordinates that make the equation $\,y = 3x + 8\,$ false.

Head up to wolframalpha.com and type in:
y = 3x + 8, x = -1, y = 5
(Cut-and-paste, if you want.)
You'll see a graph of the line, with the given point indicated by crosshairs.
you can see that going up $\,3\,$ and to the right $\,1\,$ brings you to another point on the line:
y = 3x + 8, x = -1, y = 5, x = 0, y = 8
EXAMPLE (TWO KNOWN POINTS)
Question:
Find the equation of the line through the points $\,(2,-5)\,$ and $\,(-1,4)\,$.
Write the equation in $\,y = mx + b\,$ form.
Solution:
First, use the slope formula to compute the slope:
$\displaystyle \cssId{s51}{\text{slope}} \cssId{s52}{= \frac{4-(-5)}{-1-2}} \cssId{s53}{= \frac{9}{-3}} \cssId{s54}{= -3}$
Then, continue as in the previous example:
 $y = mx + b$ (start with slope-intercept form) $y = -3x + b$ (substitute the now-known slope, $\,-3\,$, in for $\,m\,$) $4 = -3(-1) + b$ (Which point should you use? It doesn't matter! In general, try to choose the simplest numbers to work with.) $4 = 3 + b$ (simplify) $\,b = 1\,$ (subtract $\,3\,$ from both sides; write in the conventional way) $y = -3x + 1$ (substitute the now-known value of $\,b\,$ into the equation)
You might want to check that the two points do indeed lie on the line:
$-5\ \overset{\text{?}}{ = } -3(2) + 1\,$     Check!
$4\ \overset{\text{?}}{ = } -3(-1) + 1\,$     Check!
Master the ideas from this section