The point (x,-1) is 13 units from the point (3,11), What are the possible values of x ?
(x-3)^2 + (y-11)^2 = 13^2 = 169
y = -1
Solve those two equations simultaneously for x.
(x-3)^2 +(12) = 169
(x-3)^2 = 25
x - 3 = + or - 5
x = -2 or 8. y = -1 in both cases
thanks
To find the possible values of x, we need to determine the distance between two points on a coordinate plane. In this case, we know that the point (x, -1) is 13 units away from the point (3, 11).
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Using this formula, we can calculate the distance between the two points:
13 = sqrt((3 - x)^2 + (11 - (-1))^2)
Solving this equation step by step:
13^2 = (3 - x)^2 + (11 + 1)^2
169 = (3 - x)^2 + (12)^2
Taking the square root of both sides gives us:
sqrt(169) = sqrt((3 - x)^2 + (12)^2)
13 = sqrt((3 - x)^2 + 144)
Squaring both sides once again:
13^2 = ((3 - x)^2 + 144)
169 = (3 - x)^2 + 144
Expanding the equation:
169 = 9 - 6x + x^2 + 144
0 = x^2 - 6x + 144 + 9 - 169
x^2 - 6x - 16 = 0
Now we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
Applying the values to the formula:
x = (-(-6) ± sqrt((-6)^2 - 4(1)(-16))) / 2(1)
x = (6 ± sqrt(36 + 64)) / 2
x = (6 ± sqrt(100)) / 2
x = (6 ± 10) / 2
Simplifying further:
x = (6 + 10) / 2 = 16 / 2 = 8
x = (6 - 10) / 2 = -4 / 2 = -2
Therefore, the possible values of x are 8 and -2.