Given triangle DEF with vertices D(-3;6) ,E(-2;-3) and F(x;1) with DE=DF, determine the value of x
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.
To determine the value of x in triangle DEF, we can use the fact that DE = DF.
The distance between two points in a coordinate plane can be found using the distance formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the length of DE using the coordinates of points D and E:
DE = √((-2 - (-3))^2 + (-3 - 6)^2)
= √((1)^2 + (-9)^2)
= √(1 + 81)
= √82
Now, let's calculate the length of DF using the coordinates of points D and F:
DF = √((x - (-3))^2 + (1 - 6)^2)
= √((x + 3)^2 + (-5)^2)
= √(x^2 + 6x + 9 + 25)
= √(x^2 + 6x + 34)
Since DE = DF, we can equate these two distances:
√82 = √(x^2 + 6x + 34)
To get rid of the square root, we need to square both sides of the equation:
82 = x^2 + 6x + 34
Rearranging the equation:
x^2 + 6x + 34 - 82 = 0
x^2 + 6x - 48 = 0
We can now solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, we will use factoring:
(x + 8)(x - 6) = 0
Setting each factor equal to zero:
x + 8 = 0
x = -8
x - 6 = 0
x = 6
Therefore, the possible values of x in triangle DEF are x = -8 or x = 6.