Equations of perpendicular lines and points of intersection

[09jk]

hey.
I was doing some beginner yr 12 homework and overcame a problem.
question straight out of the textbook.

3 points which are A, B and C
A(1,7)
B(7,5)
C(0,-2)

textbook asks to find the equation of the perpendicular bisector of AB

I found the gradient of AB by (5-7)/(7-1)
m[sub:1dwoawon]AB[/sub:1dwoawon] = -(1/3)

Gradient of perpendicular line was found using (m[sub:1dwoawon]1[/sub:1dwoawon]xm[sub:1dwoawon]2[/sub:1dwoawon]=-1)

got m[sub:1dwoawon]2[/sub:1dwoawon]=3

so now i have the gradient of the perpendicular line, but now i have no idea how to get the y intersect for the equation.

Thanks a heap.
Jesse

 

[wjm11]

3 points which are A, B and C
A(1,7)
B(7,5)
C(0,-2)

textbook asks to find the equation of the perpendicular bisector of AB

so now i have the gradient of the perpendicular line, but now i have no idea how to get the y intersect for the equation.

Jesse,

Good job on the slope.

Since you want the perpendicular bisector, your line must pass through the midpoint of AB. Just average the x and y values of A and B to find the midpoint.

Next use the “point-slope” form of a line to finish: y – y1 = m(x – x1), where m = 3 and (x1,y1) is your midpoint.

 

[09jk]

ah ok then.

i guess i just misunderstood the question and didn't know that a "perpendicular bisector" is half way along the line.

Thanks mate
Jesse