Step-by-step explanation:
all points on the x axis have a y value of 0.
so, we need to find all points (x, 0) with a distance of 16 to (4, -8).
the distance to a point is the Hypotenuse of the right-angled triangle with the coordinate differences as legs.
and so Pythagoras applies :
c² = a² + b²
with c being the Hypotenuse (side opposite of the 90° angle).
so, in our case we have
16² = (4-x)² + (-8 - 0)²
256 = 16 - 8x + x² + 64 = x² - 8x + 80
x² - 8x - 176 = 0
if we have no other idea of factorization the general solution of such a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = -8
c = -176
x = (8 ± sqrt((-8)² - 4×1×-176))/(2×1) =
= (8 ± sqrt(64 + 704))/2 = (8 ± sqrt(768))/2 =
= (8 ± sqrt(256×3))/2 = (8 ± 16×sqrt(3))/2 =
= 4 ± 8×sqrt(3)
x1 = 4 + 8×sqrt(3) = 17.85640646...
x2 = 4 - 8×sqrt(3) = -9.856406461...
so, the points on the x axis that are 16 units apart from (4, -8) are
(x1, 0) and (x2, 0)
I am not sure in what format you need the answer.
please fill in for x1 and x2 from above what you need.
