Every non-vertical line in the coordinate plane can be described by an equation of the form
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$\,y = mx + b\,$, where:
The equation
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$\,y = mx + b\,$ 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
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$\,y = mx + b\,$ form.
Solution:
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$y = mx + b$ | (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
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$\,y = 3x + 8\,$.
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.
By adding in an additional set of 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
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$\,(2,-5)\,$ and $\,(-1,4)\,$.
Write the equation in
$\,y = mx + b\,$ form.
Solution:
First, use the
slope formula to compute the slope:
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$\displaystyle\text{slope} = \frac{4-(-5)}{-1-2} = \frac{9}{-3} = -3$
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:
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$-5\ \overset{\text{?}}{ = } -3(2) + 1\,$ Check!
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$4\ \overset{\text{?}}{ = } -3(-1) + 1\,$ Check!
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Point-Slope Form