this is a slope question

I believe that there will be two such points.

One point is 3/4ths of the way from (3,5) to (8,-3).

  To find this point:  the x-distance from 3 to 8 is 5; 3/4ths of 5 is 15/4, or 3.75, adding 3.75 onto 3 is 6.75

                             the y-distance from 5 to -3 is -8; 3/4ths of -8 is -6, adding -6 to 5 is -1

          this point is (6.75, -1)

          Checking:  the distance from 3 to 6.75 is 3.75; the distance from 6.75 to 8 is 1.25; 3.75 = 3 x 1.25

                          the distance from 5 to -1 is -6; the distance from -1 to -3 is -2; -6 = 3 x -2

The other point is past (8,-3).

   To find this point: the x-distance from 3 past 8 must have value 3, the x-distance from 8 away from 3 must                                have value 1.   Dividing the distance from 3 to 8 into 2 parts, each part has value 2.5;                                 add 2.5 onto 8 gives 10.5.

                              the y-distance from 5 to -3 has 2 parts, each part having value -4; add -4 onto -3 gives                                 -7

            this point is (10.5,-7)

          Checking:  the distance from 3 to 10.25 is 7.25; the distance from 8 to 10.25 is 2.25; 7.25 = 3 x 2.25

                          the distance from 5 to -7 is -12; the distance from -7 to -3 is -4; -12 = 3 x -4

Suppose R id (3, 5) and S is (8, -3).  Find each point on the line through R and S that is three times as far from R as it is from S

$$\vec{r}=\left(\begin{array}{r} 3 & 5 \end{array}\right) \quad \vec{s}=\left(\begin{array}{r} 8 & -3 \end{array}\right) \quad \lambda=\frac{3}{4}\\ \boxed{\vec{x}=\vec{r}+\lambda\left( \vec{s}-\vec{r}\right) \small{ \text{ line equation in vector-form}}}\\ \vec{x} =\left(\begin{array}{r} 3 & 5 \end{array}\right) +\frac{3}{4}\left( \left(\begin{array}{r} 8 & -3 \end{array}\right)- \left(\begin{array}{r} 3 & 5 \end{array}\right) \right)\\ \vec{x} =\left(\begin{array}{r} 3 & 5 \end{array}\right) +\frac{3}{4} \left(\begin{array}{r} 5 & -8 \end{array}\right) \\ \vec{x} =\left(\begin{array}{r} 3 +\frac{3}{4}* 5 & 5 +\frac{3}{4}*( -8 )\end{array}\right) \\ \vec{x} =\left(\begin{array}{r} 6.75 & -1\end{array}\right) \\$$