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**Unformatted text preview: **aining the points (−1, 3) and (2, 1).
Solution. In order to use Equation 2.2 we need to ﬁnd the slope of the line in question. So we
∆y
1
use Equation 2.1 to get m = ∆x = 2−−31) = − 2 . We are spoiled for choice for a point (x0 , y0 ).
3
(−
We’ll use (−1, 3) and leave it to the reader to check that using (2, 1) results in the same equation.
Substituting into the point-slope form of the line, we get
y = m ( x − x 0 ) + y0
2
y = − (x − (−1)) + 3
3
2
2
y = − x− +3
3
3
7
2
y = − x+ .
3
3
We can check our answer by showing that both (−1, 3) and (2, 1) are on the graph of y = − 2 x + 7
3
3
algebraically, as we did in Section 1.3.
In simplifying the equation of the line in the previous example, we produced another form of a
line, the slope-intercept form. This is the familiar y = mx + b form you have probably seen in
Intermediate Algebra. The ‘intercept’ in ‘slope-intercept’ comes from the fact that if we set x = 0,
we get y = b. In other words, the y -inte...

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