Question

How can I find an equation for the perpendicular bisector of the line segment that has the endpoints , (9,7) and (−3,−5)?

Answer 1

`y=-x+4`

Explanation:

.

`(9,7) and (-3,-5)`

First, we find the midpoint of the segment connecting these two points. The midpoint formulas are:

Midpoint `((x_1+x_2)/2, (y_1+y_2)/2)=((9-3)/2, (7-5)/2)`

Midpoint `(3,1)`

Then, we find the slope of the line segment:

`slope=m_1=((y_2-y_1)/(x_2-x_1))=((-5-7)/(-3-9))=(-12)/(-12)=1`

For a line to be perpendicular to this line, its slope `m_2` can be found from the following equation:

`m_1m_2=-1`

`(1)(m_2)=-1`

`m_2=-1`

Now, we can write the equation of the perpendicular bisector. We know its slope is `-1` and it goes through the midpoint `(3,1)`.

The equation of a straight line in slope-intercept form is:

`y=mx+b` where `m` is the slope and `b` is the `y`-intercept.

`y=m_2x+b`

`y=-x+b`

We can use the coordinates of the midpoint in this equation to solve for `b`:

`1=-3+b`

`b=4`

Therefore, the equation of the perpendicular bisector is:

`y=-x+4`

Answer 2

The midpoint of a line segment with endpoints, `(x_1, y_1)` and `(x_2,y_2)` is:

`(x_"mid", y_"mid") = ((x_1+x_2)/2, (y_1+y_2)/2)`

Substitute the given points:

`(x_"mid", y_"mid") = ((9+ -3)/2, (7+ -5)/2)`

`(x_"mid", y_"mid") = (6/2, 2/2)`

`(x_"mid", y_"mid") = (3, 1)`

Please understand that the perpendicular bisector must pass through the point `(3,1)`.

Compute the slope of the line segment:

`m = (y_1-y_2)/(x_1-x_2)`

Substitute the given points:

`m = (7--5)/(9--3)`

`m = 12/12`

`m=1`

The slope, `n`, of a line perpendicular to the line segment is:

`n = -1/m`

Substitute the value of `m`:

`n = -1/1`

`n = -1`

Use the point-slope form of the equation of a line:

`y = n(x-x_"mid")+y_"mid"`

Substitute the slope, `n`, and the midpoint:

`y = -1(x-3)+1`

We simplify and find that the equation of the perpendicular bisector is

`y = 4-x`