Given triangle DEF with vertices D(-3;4),E(-2;-3)and F(x;1) with DE=DF Determine the value of x
We can use the distance formula to find the length of DE and DF:
DE = sqrt(((-2) - (-3))^2 + ((-3) - 4)^2) = sqrt(1 + 49) = sqrt(50)
DF = sqrt((x - (-3))^2 + (1 - 4)^2) = sqrt((x + 3)^2 + 9)
Since DE = DF, we can set them equal to each other:
sqrt(50) = sqrt((x + 3)^2 + 9)
Squaring both sides gives:
50 = (x + 3)^2 + 9
Expanding and simplifying gives:
41 = x^2 + 6x
Rearranging and factoring gives:
x(x + 6) = -41
Since x cannot be negative for a point to lie to the right of D, we can discard the negative solution. So:
x + 6 = -41/x
Multiplying both sides by x:
x^2 + 6x = -41
Rearranging and factoring gives:
(x + 3)^2 = 2
Taking the square root of both sides (ignoring the negative root since x cannot be negative):
x + 3 = sqrt(2)
Solving for x:
x = sqrt(2) - 3
Therefore, the value of x is approximately -1.59.