Stitz-Zeager_College_Algebra_e-book

Incidentally this equation is sometimes called the

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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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