Question

find the equation in slope intercept form of a line that is a perpendicular bisector of segment AB with endpoints A(-5,5) and B(3,-3)

Answer 1

The equation in slope intercept form of a line that is a perpendicular bisector of segment AB with endpoints A(-5,5) and B(3,-3) is y = x + 2

Solution:

Given, two points are A(-5, 5) and B(3, -3)

We have to find the perpendicular bisector of segment AB.

Now, we know that perpendicular bisector passes through the midpoint of segment.

The formula for midpoint is:

`\text { midpoint }=\left((x_(1)+x_(2))/(2), (y_(1)+y_(2))/(2)\right)`

`Here x_1 = -5 ; y_1 = 5 ; x_2 = 3 ; y_2 = -3`

`\text { So, midpoint of } A B=\left((-5+3)/(2), (5+(-3))/(2)\right)=\left((-2)/(2), (2)/(2)\right)=(-1,1)`

Finding slope of AB:

`\text { Slope of } A B=(y_(2)-y_(1))/(x_(2)-x_(1))`

`\text { Slope } m=(-3-5)/(3-(-5))=(-8)/(8)=-1`

We know that product of slopes of perpendicular lines = -1

So, slope of AB
`*` slope of perpendicular bisector = -1

- 1
`*` slope of perpendicular bisector = -1

Slope of perpendicular bisector = 1

We know its slope is 1 and it goes through the midpoint (-1, 1)

The slope intercept form is given as:

y = mx + c

where "m" is the slope of the line and "c" is the y-intercept

Plug in "m" = 1

y = x + c ---- eqn 1

We can use the coordinates of the midpoint (-1, 1) in this equation to solve for "c" in eqn 1

1 = -1 + c

c = 2

Now substitute c = 2 in eqn 1

y = x + 2

Thus y = x + 2 is the required equation in slope intercept form