SOLUTION: Points A,B and C are 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

Question 1209204: Points A,B and C are 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 coordinates of C.
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

The set of points equidistant from two given points P and Q is the perpendicular bisector of the segment PQ. Since in this problem AC=BC, point C lies on the perpendicular bisector of AB.

So point C lies on the perpendicular bisector of segment AB and on the line AC, for which the equation is given. Point C is then the common solution to the given equation and the equation of the perpendicular bisector of AB.

I will outline the steps to the solution and let the student fill in the details.

(1) Use the given coordinates of A and B to find that the midpoint of AB is M(4,-1) and the equation of the line containing segment AB is y = -x+3.

(2) Use the coordinates of the midpoint M and the fact that the slopes of perpendicular lines are negative reciprocals to find that the equation of the perpendicular bisector is y=x-5.

(3) Use the given equation of line AC and the equation of the perpendicular bisector of AB to find that the common solution to those two equations, which is point C, is (x,y) = (50/3,35/3).

ANSWER: (50/3,35/3)

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