Given triangle DEF with vertices D(-3;6) ,E(-2;-3) and F(x;1) with DE=DF, determine the value of x

1 answer

DE = √(1^2+9^2) = √82
DF = √((x+3)^2+5^2) = √(x^2+6x+34)

So,

x^2+6x+34 = 82

check my math, and solve for x.