Question

point C is located at (1,2) and point D is located at (-4,-2). Find the x value for the point that is 1/4 the distance from point C to point D.

Answer 1

let
g(t)=(1 - 5 t, 2 - 4 t)
t between 0 and 1.
Notice that
g(0)=(1,2)=C
g(1)=(-4,-2)=D
We need to compute
g(1/5)=(0,6/5)
This is your point.
Why?

Answer 2

Let us call A=(0,6/5). Compute the distance from A and from C and compare them
\[
d_C=\frac{\sqrt{41}}{5}\\
d_D=\frac{4 \sqrt{41}}{5}\\
\frac {d_C}{d_A}=\frac 1 4
\]

Answer 3

Another way is to compute
\[
\frac 1 5 D + \frac 45 C=\frac 1 5(-4,-2) + \frac 45 (1, 2)=\\
\left(-\frac 4 5 + \frac 4 5, -\frac 2 5+ \frac 85
\right)=\left(0, \frac 6 5

\right)
\]