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?

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

Looks pretty good to me!