Review question

# Can we find the point that divides a line in a given ratio? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9585

## Solution

The point lying between $P(2,3)$ and $Q(8,-3)$ which divides the line $PQ$ in the ratio $1:2$ has co-ordinates

1. $(4,-1)$

2. $(6,-2)$

3. $(\frac{14}{3}, 2)$

4. $(4,1)$

#### Approach 1: Similar triangles

We sketch the situation:

Let $R = (a,b)$ divide the line $PQ$ in the ratio $1:2$ with $PR:RQ = 1:2$, so $R$ lies one third of the way along $PQ$.

By similar triangles, $\dfrac{PR}{PQ} = \dfrac{1}{3} = \dfrac{a-2}{6}=\dfrac{3-b}{6}$, which gives $a = 4$, $b =1$.

#### Approach 2: Vectors

We can also answer this question using vectors.

Now $\vec{OP} = \begin{pmatrix}2\\3\end{pmatrix}$, while $\vec{OQ} = \begin{pmatrix}8\\-3\end{pmatrix}$.

Hence $\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix}6\\-6\end{pmatrix}$.

Now $\vec{OR}= \vec{OP} + \frac{1}{3}\vec{PQ} = \begin{pmatrix}2\\3\end{pmatrix}$ + $\begin{pmatrix}2\\-2\end{pmatrix}=\begin{pmatrix}4\\1\end{pmatrix}$.

Either way, we reach the answer d.