Alg6.2

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

(12,1),(4,0)\,

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

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

To find the (always zero or positive) distance, use the formula d=+{\sqrt  {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}\,

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

To find the slope, use the formula m={\frac  {y_{2}-y_{1}}{x_{2}-x_{1}}}\,

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

The equations of the line are

Form 1: y=mx+b\,

Plug in one known point (say, (4,0)\,) and the calculated slope.

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

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

Now plug b and m into the line equation:

  • y={\frac  {1}{8}}x-{\frac  {1}{2}}\,

Form 2: (y-y_{1})=m(x-x_{1})\,

Plug in one known point (say, (12,1)\,) and the calculated slope.

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

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

  • y={\frac  {1}{8}}x-{\frac  {1}{2}}\,

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