Find X such that the point (X, 8) is 13 units from (7, -4)
√((x-7)^2 + 12^2) = 13
square both sides and expand
x^2 - 14x + 49 + 144 = 169
x^2 - 14x +24 = 0
(x-12)(x-2) = 0
x = 12 or x = 2
To find the value of X such that the point (X, 8) is 13 units from (7, -4), we can use the distance formula in the Cartesian plane.
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, the given point is (7, -4), and the other point is (X, 8).
Substituting the known values into the distance formula, we get:
13 = sqrt((X - 7)^2 + (8 - (-4))^2)
Simplifying further:
169 = (X - 7)^2 + 12^2
169 = (X - 7)^2 + 144
Now, let's solve for X.
Move the constant term to the other side of the equation:
(X - 7)^2 = 169 - 144
(X - 7)^2 = 25
Taking the square root of both sides:
X - 7 = ± sqrt(25)
X - 7 = ±5
Now, we have two possible values for X:
1. X - 7 = 5
Add 7 to both sides:
X = 5 + 7
X = 12
2. X - 7 = -5
Add 7 to both sides:
X = -5 + 7
X = 2
Therefore, the two possible values of X are 12 and 2.