Suppose $R$ is $(3, 5)$ 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 $S$.

Question

Suppose $R$ is $(3, 5)$ 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 $S$.

I'm confused regarding "three times as far from $R$ as it is from $S$", like is it equally $3$ times far from $R$ being equidistant and also $3$ times from $S$ since I'm not familiar with these question but I think I've found one point.

Let $P$ be the point on segment $\overline{RS}$ such that $RP:PS=3:1$

$\therefore \frac{RP}{RS}=\frac{3}{4}$

$ \frac{3}{4}RS=RP$

Since S is $5$ units to the right and $8$ units below $R$

$P$ is $5(\frac{3}{4})$ units to the right and $8(\frac{3}{4})$ units below $R$

$\therefore P:(3+\frac{15}{4},5-\frac{24}{4})=(6.75,-1)$

I need help with the other one, thanks in advance.

Answer 1

If $T$ is that point then you have $$ST = 3RT\implies ||\vec{ST}||= 3||\vec{RT}|| $$

So you have two cases:

• $\vec{ST}= 3\vec{RT} \implies \vec{T} = {1\over 2}(3\vec{R}-\vec{S})$
• $\vec{ST}= -3\vec{RT}\implies \vec{T} = {1\over 4}(3\vec{R}+\vec{S})$