f (1, -2) is equidistant with (4, 4) and (x, 4), find x.
This would have been a little tougher had the two points not lain on a horizontal line.
As it is, (1,-2) lies 3 units to the left of (4,4)
So, just pick the point (x,4) such that 1 is 3 units to the right of x.
x = -2
f (1, -2) is equidistant with (4, 4) and (x, 4), find x.
I need a complete solution for this equation.
To find the value of x in the equation, we need to determine the distance between the point (1, -2) and the two given points: (4, 4) and (x, 4). Since the point (1, -2) is equidistant from these two points, we know that the distances from (1, -2) to (4, 4) and (1, -2) to (x, 4) are equal.
Let's calculate these distances step by step:
Distance between (1, -2) and (4, 4):
Using the distance formula, which is √((x2 - x1)^2 + (y2 - y1)^2), we can substitute the values:
√((4 - 1)^2 + (4 - (-2))^2)
Simplifying this:
√(3^2 + 6^2)
√(9 + 36)
√45
Distance between (1, -2) and (x, 4):
Again, using the distance formula:
√((x - 1)^2 + (4 - (-2))^2)
Simplifying:
√((x - 1)^2 + 6^2)
√((x - 1)^2 + 36)
√(x^2 - 2x + 1 + 36)
√(x^2 - 2x + 37)
Since we know the distances are equal, we can equate them and solve for x:
√45 = √(x^2 - 2x + 37)
By squaring both sides of the equation to eliminate the square roots:
45 = x^2 - 2x + 37
Rearranging terms:
x^2 - 2x - 8 = 0
To solve this quadratic equation, we can factor or use the quadratic formula. Factoring is not possible in this case, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In our case, a = 1, b = -2, and c = -8. Plugging in these values:
x = (-(-2) ± √((-2)^2 - 4(1)(-8))) / (2(1))
x = (2 ± √(4 + 32)) / 2
x = (2 ± √36) / 2
x = (2 ± 6) / 2
This gives us two possible solutions:
x₁ = (2 + 6) / 2 = 8 / 2 = 4
x₂ = (2 - 6) / 2 = -4 / 2 = -2
Thus, the values of x that make (1, -2) equidistant from (4, 4) and (x, 4) are x = 4 and x = -2.