1. ## Perpendicular bisector

The main help i need is with this question:

1)****The perpendicular bisector of a straight line joining the points (3,2) and (5,6) meets the x axis at A and the y at B. Prove that the distance AB is equal to 6root5.

I just need to have theses checked to make sure there right, would love if you could just skim it and check

1)What is the gradient of the line perpendicular to y=3x-5. hence find the equation of the line which goes through (3,1) and is perpendicular to y=3x-5
a) to find the gradient I did, m2=-1/m1. m1=3 so it was -1/3 = -1/3
b) We know the coordinates (3,1) which is the y and X, so 1=(-1/3*3)+c
1=-1+c, so
2=C all together the answer is y=-1/3x+2

2)Find the equation of line B which is perpendicular to line A and goes though the mid-point.
I am given the coordinates for line A, (2,6) and (4,8). Using that i can find both the midpoint and the gradient of line A = midpoint = (3,7) Gradient = 1
So M2=1/M1 so -1/1=-1
we know from midpoint, y=7 and x=3 so
7=(-1*3)+c
7+3=c
10=c so the equation is Y=-x+10

2. Originally Posted by Greener
The main help i need is with this question:
1)****The perpendicular bisector of a straight line joining the points (3,2) and (5,6) meets the x axis at A and the y at B. Prove that the distance AB is equal to 6root5.
There is a lot wrong with that.
Start with the midpoint between $(3,2)~\&~(5,6)$ is $(4,4)$.

3. ## perpendicular bisector

Hi greener,
Did Plato's hint help you solve this problem? My guess is that you may still be struggling with it.If so here are a few suggestions. Plot the two given points Connect them Draw a slope diagram. From that calculate the slope of the line between the points .
What then is the slope of the perpendicular bisector It passes thru the midpoint 4,4.Use the point slope formula to find the equation.I hope you can handle the rest.

bjh