Given triangle DEF with vertices D(-3;4),E(-2;-3)and F(x;1) with DE=DF
Determine the value of x
First, we need to find the distance between points D and E, and between points D and F, to see if DE is equal to DF:
- Distance between D and E:
dDE = sqrt[(x2-x1)^2 + (y2-y1)^2]
dDE = sqrt[(-2+3)^2 + (-3-4)^2]
dDE = sqrt(1 + 49)
dDE = sqrt(50)
- Distance between D and F:
dDF = sqrt[(x2-x1)^2 + (y2-y1)^2]
dDF = sqrt[(x+3)^2 + (1-4)^2]
dDF = sqrt[(x+3)^2 + 9]
Since it's given that DE is equal to DF, we can equate their distances:
sqrt(50) = sqrt[(x+3)^2 + 9]
Squaring both sides, we get:
50 = (x+3)^2 + 9
41 = (x+3)^2
Taking the square root of both sides (with a positive root since distance cannot be negative), we get:
x+3 = sqrt(41)
x = sqrt(41) - 3
Therefore, the value of x is approximately -0.146.