Question

the longer base of a trapezoid has endpoints of (-2,-4) and (4,0). the shorter base contains the point (3,1). write an equation of the line that contains the shorter base of the trapezoid.

Explanation:

Step1: Find slope of longer - base

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For the points $(-2,-4)$ and $(4,0)$ of the longer - base, $m_1=\frac{0-(-4)}{4 - (-2)}=\frac{4}{6}=\frac{2}{3}$. Since the bases of a trapezoid are parallel, the slope of the line containing the shorter base is the same as the slope of the line containing the longer base, so $m = \frac{2}{3}$.

Step2: Use point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$. We have the point $(3,1)$ and $m=\frac{2}{3}$. Substitute $x_1 = 3$, $y_1 = 1$, and $m=\frac{2}{3}$ into the point - slope form: $y - 1=\frac{2}{3}(x - 3)$.

Step3: Convert to slope - intercept form

Expand the right - hand side: $y - 1=\frac{2}{3}x-2$. Then add 1 to both sides to get $y=\frac{2}{3}x-1$.

Answer:

$y=\frac{2}{3}x - 1$