Question

The vertices of an isosceles triangle are A(3, 6), B(7, 2), and C(4, 3).

What is the equation of the triangle's line of symmetry?

Answer 1

I belive it is y = - x + 1

Answer 2

Check first which sides are equal in length easy to see this using pythagorus theorem. distance between A and B is sqr root of (7-3)^2 +(2-6)^2 = sqr root 32 distance between B and C is sqr root of (4-7)^2 +(3-2)^2 = sqr root 10 distance between C and A is sqr root of (4-3)^2 + (3-6)^2= sqr root 10 so BC = CA so the line of symmetry must go through the other side AB centre of AB is [(7+3)/2, (2+6)/2] = (5,4) and the line must go through the opposite vertex C so finding the line through the 2 points (5,4) and (4,3) y=mx+c m= (4-3)/5-4) =1 y=x+c substitute either point (4,3) gives 3=4+c c=-1 so equation is y=x-1 Check sum: substitute the other point (5,4) 4=5-1 yes so all correct