SOLUTION: Points A, B and C are the three vertices of an isosceles triangle in which AC=BC. The coordinates of A and B are (-2,1) and (6,-3) respectively and the equation of AC is 4x-7y+15=0

Question 1195338: Points A, B and C are the three vertices of an isosceles triangle in which AC=BC. The coordinates of A and B are (-2,1) and (6,-3) respectively and the equation of AC is 4x-7y+15=0. Find the value of the coordinates of C
Found 2 solutions by amarjeeth123, greenestamps:
Answer by amarjeeth123(569) (Show Source): You can put this solution on YOUR website!
Points A, B and C are the three vertices of an isosceles triangle in which AC=BC. The coordinates of A and B are (-2,1) and (6,-3) respectively and the equation of AC is 4x-7y+15=0.
Let the coordinates of C be (h,k).
We are given that AC=BC
Squaring both sides we get AC^2=BC^2
The distance between two points(x1,y1) and (x2,y2) is given by sqrt((y2-y1)^2+(x2-x1)^2)
Plugging in the values we get, AC^2=(h+2)^2+(k-1)^2=h^2+4h+4+k^2-2k+1=h^2+k^2+4h-2k+5
We also have BC^2=(h-6)^2+(k+3)^2=h^2-12h+36+k^2+6k+9=h^2+k^2-12h+6k+45
Equating both we get,
h^2+k^2+4h-2k+5=h^2+k^2-12h+6k+45
4h-2k+5=-12h+6k+45
16h-8k=40
Dividing by 8 on both sides we get,
2h-k=5................................Equation 1
Since (h,k) lies on the line AC we get,
4h-7k+15=0----------------------------Equation 2
multiplying Equation 1 by 7 we get,
14h-7k=35.............................Equation 3
Subtracting equation 2 from 3 we get,
14h-7k-4h+7k-15=35
Simplifying we get,
10h=50
h=5
Substituting in equation 1 we get,
10-k=5
k=5
The point C has coordinates (5,5)
AC^2=49+16=65 square units
BC^2=1=64=65 square units
The point (5,5) also satisfies the equation 4x-7y+15=0
Hence C(5,5) is the correct solution.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Because AC=BC, point C is on the perpendicular bisector of segment AB. So find the equation of that perpendicular bisector and then solve the pair of linear equations to find the coordinates of C.

AB has slope (-3-1)/(6-(-2)) = -1/2 and midpoint (2,-1).

The perpendicular bisector has slope 2 and passes through (2,-1); its equation is y=2x-5, or 2x-y-5 = 0. So we have

(1) 4x-7y+15 = 0
(2) 2x-y-5 = 0

Double the second equation and subtract one equation from the other to find y:

4x-7y+15 = 0
4x-2y-10 = 0

-5y+25 = 0
5y = 25
y = 5

Substitute y=5 in either equation to find x:

2x-5-5 = 0
2x-10 = 0
2x = 10
x = 5

ANSWER: C is (5,5)

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