Asked by Anonymous
Given triangle DEF with vertices D(-3;6) ,E(-2;-3) and F(x;1) with DE=DF, determine the value of x
Answers
Answered by
Steve
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.
DF = √((x+3)^2+5^2) = √(x^2+6x+34)
So,
x^2+6x+34 = 82
check my math, and solve for x.
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