# Difference between revisions of "Alg6.2"

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

\n\n$\displaystyle (12,1),(4,0)\,$

\n\nTo find the midpoint, average the x coordinates and y coordinates. The midpoint is

\n\n$\displaystyle \left(\frac{12+4}{2},\frac{1+0}{2}\right) = \left(8,\frac{1}{2}\right)\,$

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

\n\n$\displaystyle d = \sqrt{(12-4)^2+(1-0)^2} = \sqrt{(8)^2+1^2} = \sqrt{64+1} = \sqrt{5*15} = \sqrt{5*3*5} = 5\sqrt{3}\,$

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

\n\n$\displaystyle m = \frac{0-1}{4-12} = \frac{-1}{-8} = \frac{1}{8}\,$

\n\nThe equations of the line are

\n\nForm 1: $\displaystyle y=mx+b\,$

\n\nPlug in one known point (say, $\displaystyle (4,0)\,$ ) and the calculated slope.

\n\n$\displaystyle 0 = \frac{1}{8}\cdot 4 + b\,$

\n\n$\displaystyle b = -\frac{4}{8} = -\frac{1}{2}\,$

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

\n\n*$\displaystyle y = \frac{1}{8}x - \frac{1}{2}\,$

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

\n\nPlug in one known point (say, $\displaystyle (12,1)\,$ ) and the calculated slope.

\n\n$\displaystyle (y-1) = \frac{1}{8}(x-12)\,$

\n\n$\displaystyle y = \frac{1}{8}x - \frac{12}{8} + 1 = \frac{1}{8}x - \frac{4}{8} \,$

\n\n*$\displaystyle y = \frac{1}{8}x - \frac{1}{2}\,$