Given (12, 1) and (4, 0), find the midpoint, distance, slope, and equation of the line.

Problem: Given these pairs of points, (12, 1) and (4, 0), find the midpoint, distance, slope, and equation of the line.

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

Solutions:

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

Method 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:

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

Method 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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Tags: distance, equation of line, Geometry Problems, midpoint, slope

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Saturday, December 28, 2013