# How do I find the equation of a perpendicular bisector of a line segment with the endpoints (-2, -4) and (6, 4)?

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Jan 23, 2016

#### Answer:

$x + y - 2 = 0$

#### Explanation:

Here's how.

You need to find the slope of the segment since it is the negative reciprocal of the slope of the bisector.
Another one you will have to find is the midpoint of the segment because the bisector will pass trough that point.

If you have those two, you can now determine the equation of the perpendicular bisector using point-slope form of a line.

Let:
${m}_{s} \implies$slope of the segment
${m}_{b} \implies$slope of the bisector
$\left({x}_{1} , {y}_{1}\right) \implies \left(- 2 , - 4\right)$
$\left({x}_{2} , {y}_{2}\right) \implies \left(6 , 4\right)$
$M \left({x}_{m} , {y}_{m}\right) \implies$midpoint of segment

Solving for the slope of the segment:

${m}_{s} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{4 - \left(- 4\right)}{6 - \left(- 2\right)} = \frac{8}{8} = 1$

Solving for the slope of the bisector:

${m}_{b} = - \frac{1}{m} _ s = - \frac{1}{1} = - 1$

Solving for the midpoint of the segment:

${x}_{m} = \frac{{x}_{1} + {x}_{2}}{2} = \frac{- 2 + 6}{2} = 2$

${y}_{m} = \frac{{y}_{1} + {y}_{2}}{2} = \frac{4 - 4}{2} = 0$

$M \left(2 , 0\right)$

Solving for the equation of the segment:

POINT-SLOPE FORM

$y - {y}_{m} = {m}_{b} \left(x - {x}_{m}\right)$

$y - 0 = - 1 \left(x - 2\right)$

$y = - x + 2 \text{ } \leftarrow$slope-intercept form

$x + y - 2 = 0 \text{ } \leftarrow$ standard form

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Jul 16, 2016

#### Answer:

$x + y - 2 = 0$

#### Explanation:

Let $\left(x , y\right)$ be any point on the perpendicular bisector. From elementary geometry, we can easily see that this point must be equidistant from the two points $\left(- 2 , - 4\right) \mathmr{and} \left(6 , 4\right)$. Using the Euclidean distance formula gives us the equation

${\left(x + 2\right)}^{2} + {\left(y + 4\right)}^{2} = {\left(x - 6\right)}^{2} + {\left(y - 4\right)}^{2}$

This can be rewritten as

${\left(x + 2\right)}^{2} - {\left(x - 6\right)}^{2} = {\left(y - 4\right)}^{2} - {\left(y + 4\right)}^{2}$

Using ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, this simplifies to

$8 \left(2 x - 4\right) = - 8 \cdot 2 y$ which simplifies to

$x + y - 2 = 0$

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