Question

Please help !! My friend has a math test tomorrow and he is struggling to answer this:

Answer 1

Answer:

So the line that is a perpendicular bisector of AB is $y=\frac{3}{8}x+\frac{11}{16}$.

Step-by-step explanation:

If we are looking for a line such that is is perpendicular bisector of AB, then we first need to find the midpoint of AB.

The midpoint of AB is going to be the "average" point.  

What I mean by this, to find the x-coordinate of the midpoint we need to average our x-coordinates of our endpoints and we also need to do the same for the y-coordinate part.

Let's begin this.

The average of the x-coordinates of the endpoints are:

$\frac{2+5}{2}=\frac{7}{2}$

The average of the y-coordinates of the endpoints are:

$\frac{6+(-2)}{2}=\frac{4}{2}=2$

This concludes the midpoint of AB is $(\frac{7}{2},2)$.

Now we also need to calculate the slope of $AB$ so we can find the slope of the line perpendicular to it.

To do this we could use $\frac{y_2-y_1}{x_2-x_1}$. Without using the formula directly, I like to line the points up vertically and subtract. I then put the 2nd difference over the first difference. Like so,

  $(5,-2)$

-$(2,6)$

----------------------

$3 , -8$

So the slope is $\frac{-8}{3}$.

The line that is perpendicular to the line AB will have an opposite reciprocal slope.

So the line we are looking for has slope $\frac{3}{8}$.

So we are looking for a line going through the midpoint $(\frac{7}{2},2)$ and has slope $\frac{3}{8}$.

A line in slope-intercept form is $y=mx+b$.

We know $m=\frac{3}{8}{/tex] and we know a point [tex](x,y)=(\frac{7}{2},2)$. We could plug this in to find $b$.

Let's do that now.

$2=\frac{3}{8}(\frac{7}{2})+b$

Simplify:

$2=\frac{21}{16}+b$

Subtract 21/16 on both sides:

$2-\frac{21}{16}=b$

Find a common denominator:

$\frac{32}{16}-\frac{21}{16}$

$\frac{11}{16}$

So the line that is a perpendicular bisector of AB is $y=\frac{3}{8}x+\frac{11}{16}$.  This is so because the line we found will be perpendicular to AB while also cutting AB into two equal halves.