# Alg6.1

For these pairs of points, find the midpoint, distance, slope, and equation of the line.

$\displaystyle (-2,10),(4,9)\,$

To find the midpoint, average the x coordinates and y coordinates. The midpoint is

$\displaystyle \left(\frac{-2+4}{2},\frac{10+9}{2}\right) = \left(\frac{2}{2},\frac{19}{2}\right) = \left(1,\frac{19}{2}\right)\,$

To find the (always zero or positive) distance, use the formula $\displaystyle d = +\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\,$

$\displaystyle d = \sqrt{(-2-4)^2+(10-9)^2} = \sqrt{(-6)^2+1^2} = \sqrt{36+1} = \sqrt{37}\,$

To find the slope, use the formula $\displaystyle m = \frac{y_2-y_1}{x_2-x_1}\,$

$\displaystyle m = \frac{9-10}{4-(-2)} = \frac{-1}{6}\,$

The equations of the line are

Form 1: $\displaystyle y=mx+b\,$

Plug in one known point (say, $\displaystyle (4,9)\,$ ) and the calculated slope.

$\displaystyle 9 = \frac{-1}{6}\cdot 4 + b\,$

$\displaystyle b = 9+\frac{2}{3} = \frac{29}{3}\,$

Now plug $\displaystyle b$ and $\displaystyle m$ into the line equation:

• $\displaystyle y = \frac{-1}{6}x + \frac{29}{3}\,$

Form 2: $\displaystyle (y-y_1) = m(x-x_1)\,$

Plug in one known point (say, $\displaystyle (-2,10)\,$ ) and the calculated slope.

$\displaystyle (y-10) = \frac{-1}{6}(x-(-2))\,$

$\displaystyle y = \frac{-1}{6}x - \frac{2}{6} + 10 = \frac{-1}{6}x + \frac{58}{6} \,$

• $\displaystyle y = \frac{-1}{6}x + \frac{29}{3}\,$