# Finding triangle coordinates

Alignments to Content Standards: G-GPE.B.6 G-SRT.B.5

Below is a picture of a triangle $ABC$ on the coordinate grid. The red lines are parallel to $\overleftrightarrow{BC}$:

Suppose $P = (1.2,1.6)$, $Q = (2,4)$, and $R = (2.4,5.2)$. Find the coordinates of the points $u$, $v$, and $w$.

## IM Commentary

The purpose of this task is to use similar triangles in order to study the coordinates of points which divide a line segment in a given ratio. These coordinates can be calculated directly but the method employed here not only allows us to find the coordinates of the desired points but also to construct them (with straightedge and compass). Indeed, the point $B$ on our triangle can be chosen conveniently so that $\overline{AB}$ can be readily divided in the desired ratio. We can then construct the red lines parallel to $\overleftrightarrow{BC}$ and, using the result of this task, we have divided the segment $\overline{AC}$ in the desired ratio.

In order to complete this task, students will need to understand similar triangles and the proportional relationships to which they give rise. The coordinates of the points $u, v, w$, dividing $\overline{AC}$ in specific ratios, can also be found directly.

## Solution

We can calculate the coordinates of $u, v,$ and $w$ using similar triangles. We begin with $v$ as the numbers are simplest. We are given $A = (1,1)$, $B = (3,7)$, and $Q = (2,4)$. So $Q$ is the midpoint of $\overline{AB}$. Since $\overleftrightarrow{Qv}$ is parallel to $\overleftrightarrow{BC}$, we know using alternate interior and vertical angles that $m(\angle AQv) = m(\angle ABC)$ and $m(\angle AvQ) = m(\angle ACB)$. The triangles $ABC$ and $AQv$ share angle $A$ so they are similar. Therefore $$\frac{|AB|}{|AC|} = \frac{|AQ|}{|Av|}.$$ Since $Q$ is the midpoint of $\overline{AB}$ we know that $\frac{|AB|}{|AQ|} = 2$ or $\frac{|AQ|}{|AB|} = \frac{1}{2}$. Solving for $|Av|$ in the displayed equation above gives $|Av| = \frac{|AC|}{2}$, that is $v$ is the midpoint of $\overline{AC}$. Since $A = (1,1)$ and $C = (8,2)$ this means that $v = (4.5,1.5)$.

The same method can be employed for the points $P$ and $R$. Namely, $\triangle APu$ and $\triangle ARw$ are both similar to $\triangle ABC$. We therefore obtain proportions as above:

\begin{align} \frac{|AB|}{|AC|} &= \frac{|AP|}{|Au|}\\ \frac{|AB|}{|AC|} &= \frac{|AR|}{|Aw|}. \end{align}

We have $\frac{|AB|}{|AP|} = 10$ and $\frac{|AB|}{|AR|} = \frac{10}{7}$. This means that $u$ is $\frac{1}{10}$ of the way from $A$ to $C$, that is $|Au| = \frac{|AC|}{10}$ and similarly $|Aw| = \frac{7|AC|}{10}.$ From $A$ to $C$ the $x$-coordinate increases by 7 and the $y$-coordinate increases by 1. This means that

\begin{align} u &= (1.7,1.1),\\ w &= (5.9,1.7). \end{align}