NEED HELP ASAP!!!

Write the equation of the perpendicular bisector of \overline{AB} A B ¯ if A(–6, –4) and B(2, 0).
Group of answer choices
y=-2x-2 y = − 2 x − 2
This answer is set as correct T h i s a n s w e r i s s e t a s c o r r e c t
y=\frac{1}{2}x-6 y = 1 2 x − 6
y=\frac{1}{2}x-2
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NEED HELP ASAP!!! Write the equation of the perpendicular bisector of \overline{AB} A B ¯ if A(–6, –4) and B(2, 0). Group of answer choices y=-2x-2 y = − 2 x − class=

Answer :

The equation of perpendicular bisector of A(-6, -4 ) and B(2, 0) is y = -2x - 6

Solution:

Given that we have to find the equation of perpendicular bisector of A(-6, -4 ) and B(2, 0)

A perpendicular bisector, bisects a line segment at  right angles

To obtain the equation we require slope and a point on it

Find the midpoint and slope of the given points and then we can find the equation

Find the midpoint:

Given points are A(-6, -4 ) and B(2, 0)

The midpoint is given as:

[tex]m(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)[/tex]

[tex]\text {Here } x_{1}=-6 ; y_{1}=-4 ; x_{2}=2 ; y_{2}=0[/tex]

Substituting the values we get,

[tex]\begin{aligned}&m(x, y)=\left(\frac{-6+2}{2}, \frac{-4+0}{2}\right)\\\\&m(x, y)=(-2,-2)\end{aligned}[/tex]

Find the slope of given points:

[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]m=\frac{0-(-4)}{2-(-6)}\\\\m = \frac{4}{8}\\\\m = \frac{1}{2}[/tex]

Then the slope of perpendicular bisector is given as:

We know that product of slopes of given line and slope of line perpendicular to it is equal to -1

Let the slope of perpendicular bisector be [tex]m_1[/tex]

[tex]\frac{1}{2} \times m_1 = -1\\\\m_1 = -2[/tex]

Find the equation of line with slope -2 and point (-2, -2)

The equation of line in slope intercept form is given as:

y = mx + c -------- eqn 1

Where "m" is the slope and "c" is the y - intercept

Substitute (x, y) = (-2, -2) and slope m = -2 in eqn 1

-2 = -2(-2) + c

-2 = 4 + c

c = -2 - 4

c = -6

Substitute c = -6 and m = -2 in eqn 1

y = -2x - 6

Thus the required equation of perpendicular bisector is found

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