Expert: Sherman D. - 8/26/2007

Question
Hi there, i wonder if you can help me.

Im currently studying maths in school; im from scotland so im unsure whether your aware of the education system.  Im in my 5th year of high school, and studying Higher Mathematics.

In regards to my question, the class has recently begun the new course.  First up was the straight line, which i was fine with at standard grade (the two years of study that preceed higher) but at the higher level i find it has got alot harder.

I am confused by the midpoint of a line and how to find it,as is the same with the distance formula, perpendicular  bisectors and that kind of idea.

Answer
Using (2,3) and (4,5) as your 2 main coordinates.

M = ((4 + 2)/2),((5 + 3)/2)
M = (6/2),(8/2)
M = (3,4)

info found at www.purplemath.com/modules/midpoint.htm

Here's your proof

(3,4) and (2,3), (3,4) and (4,5)
sqrt((2 - 3)^2 + (3 - 4)^2) = sqrt((5 - 4)^2 + (4 - 3)^2)
sqrt((-1)^2 + (-1)^2) = sqrt(1^2 + 1^2)
sqrt(1 + 1) = sqrt(1 + 1)
sqrt(2) = sqrt(2)
2 = 2

info found at www.purplemath.com/modules/distform.htm

so since the left side equals the right side, (3,4) is your midpoint.

To find the perpendicular bisector

(2,3) and (3,4)
m = (4 - 3)/(3 - 2)
m = 1/1
m = 1

for it to be a perpendicular bisector, it must be an opposite reciprocal of the slope and it must contain the midpoint.

(3,4), m = -1
4 = -1(3) + b
4 = -3 + b
b = 7

so the perpendicular bisector equation is y = -x + 7