A problem on triangle and it's perpendicular bisectors.

[agoogler]

Homework Statement

I'm trying to solve the following problem :

In △ABC, coordinates of B are (−3,3). Equation of the perpendicular bisector of side AB is 2x+y−7=0. Equation of the perpendicular bisector of side BC is 3x−y−3=0. Mid point of side AC is E(11/2,7/2). Find AC².

Homework Equations

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The Attempt at a Solution

By solving 3x−y−3=0 and 2x+y−7=0 I find that the intersection of perpendicular bisectors is at (2,3) .
Then using the two points (2,3) and (11/2,7/2), I get the equation of perpendicular bisector of AC as y=x/7+19/7.
So the slope of AC is -7 and then using point slope form , y−7/2=−7(x−11/2) Thus the equation of line AC is y=42−7x .
Similarly equation of line BC is y=2−x/3 .
So AC and BC intersect at (6,0).
By using the fact that E is the midpoint of AC, I find Co-ordinates of A as (5,7).
So the distance between A and C is 5√2, and AC²=50.
But this answer is wrong and the correct answer is 74 ( I checked the answer sheet) .
What have I done wrong ?

 

[verty]

I don't know what went wrong before but here is a slightly different method.

Step 1. Find A using B and the gradient of AB.
Step 2. Find AC^2

I forget about E, E means only A is required.

 

[agoogler]

I don't know what went wrong before but here is a slightly different method.

Step 1. Find A using B and the gradient of AB.
Step 2. Find AC^2

I forget about E, E means only A is required.

I don't understand. How can I find AC² after finding gradient of AB ?

 

[verty]

I don't understand. How can I find AC² after finding gradient of AB ?

Your goal is to find A and then AC^2. To find A, find the midpoint of AB then use the midpoint formula.

 

[haruspex]

So the distance between A and C is 5√2, and AC²=50.

I agree with your answer, through a slightly different route.
O is the circumcentre, so OC=OB = 5. OE² = 25/2, so CE² = 25/2 by Pythagoras.

 

[agoogler]

Your goal is to find A and then AC^2. To find A, find the midpoint of AB then use the midpoint formula.

I tried and got A as (5,7) . Is that right?
Then I get C as (6,0) so AC²=50.

 

[verty]

I tried and got A as (5,7) . Is that right?
Then I get C as (6,0) so AC²=50.

It must be right, you found the same answer by two different methods and Haruspex found it by a third method. There must have been no mistake originally.