How do you find the equation in slope-intercept form of the line that is perpendicular to AB and passes through the midpoint of AB. Let A = (-6,2) and B = (4,-10)?

Apr 9, 2017

$y = \frac{5}{6} x - \frac{19}{6}$

Explanation:

The mid point is the mean value as you read left to right on the x-axis.

Point A $\to {P}_{a} \to \left({x}_{a} , {y}_{a}\right) = \left(- 6 , 2\right)$
Point B $\to {P}_{b} \to \left({x}_{b} , {y}_{b}\right) = \left(4 , - 10\right)$

Let the mean point be ${P}_{m} \to \left({x}_{m} , {y}_{m}\right)$
Let the gradient of the line between ${P}_{a} \mathmr{and} {P}_{b} \text{ be } m$

Note that the gradient of the perpendicular line will be $- \frac{1}{m}$

We read from ${x}_{a} \text{ to } {x}_{b}$

$\textcolor{b l u e}{\text{Determine the mid point}}$

${x}_{m} = \frac{{x}_{b} + {x}_{a}}{2} = \frac{4 + \left(- 6\right)}{2} = - 1$

${y}_{m} \to \frac{{y}_{b} + {y}_{a}}{2} = \frac{2 + \left(- 10\right)}{2} = - 4$

${P}_{m} \to \left({x}_{m} , {y}_{m}\right) = \left(- 1 , - 4\right)$
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$\textcolor{b l u e}{\text{Determine the gradient (slope) of the perpendicular line}}$

$m = \left(\text{change in y")/("change in x}\right) \to \frac{{y}_{b} - {y}_{a}}{{x}_{b} - {x}_{a}} = \frac{- 10 - 2}{4 - \left(- 6\right)} = \frac{- 12}{10} = - \frac{6}{5}$

Thus $- \frac{1}{m} = + \frac{5}{6}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the equation of the perpendicular line}}$
Found that $y = - \frac{1}{m} x + c$ is:

$y = \frac{5}{6} x + c \text{ } \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

We know that this passes through the point
${P}_{m} \to \left({x}_{m} , {y}_{m}\right) = \left(- 1 , - 4\right)$

So by substitution into $E q u a t i o n \left(1\right)$ we have:

$- 4 = \frac{5}{6} \left(- 1\right) + c$

$c = - 4 + \frac{5}{6} = - 3 \frac{1}{6} = - \frac{19}{6}$ giving:

$y = \frac{5}{6} x - \frac{19}{6}$ 