write an equation of the perpendicular bisector in line segment (-3,7) to (11,9)?

Question

This is how i answered the problem: m= y^2 - y^1/x^2 - x!=9-7/11-(-3)=2/13.

Answer 1

Hi Rosalind,

 

Remember that an equation of a line can be expressed as y = mx + b.

 

You have plugged your points in rightly to find the slope of the line segment, although do be careful with your subtraction. 11-(-3) = 14 so the slope of the line segment is 2/14 or 1/7.

 

We also need to find the mid-point of the line segment in order to bisect the line. To find the midpoint, you need to average the coordinates. (-3 + 11)/2 = 4 and (7 + 9) /2 = 8. This tells us that we want our bisector to pass through the point (4, 8)

 

To find the perpendicular line, we need to use the negative reciprocal of our existing slope. The negative reciprocal of 1/7 is -7.

 

We can now plug in the values using 2 point form:

y - 8 = -7(x-4)

y - 8 = -7x + 28      (distribute the -7)

y = -7x + 36           (add 8 to both sides)

 

I hope this helped!

Lindsey

Answer 2

Rosalind

 

We have a line segment connecting two points (-3,7) and (11,9)

 

You have the right start (but with an arithmetic error); you just need to work through the rest of the problem.

 

The slope of this line is

(9-7)/[11-(-3)] = 2/14 = 1/7

 

The perpendicular bisector will have a slope of -7

since lines that are perpendicular have slopes that are negative inverses of one another.

 

Now since this is a perpendicular BISECTOR, we need to find the midpoint of the points (-3,7) and (11,9)

   -3+11       9+7

(---------  , -------- )

       2            2

 

which reduces to the point

(4, 8)

 

So we need to find the equation of a line with a slope of -7 that goes through the point (4/8)

y-8 = (-7)(x-4)

y-8 = -7x+28

y = -7x+36