8.04 Polygons in the coordinate plane

Many geometrical properties of figures can either be verified or proved using coordinate geometry.

There is a range of established formulas that become useful in this endeavor. You may wish to go back to a previous lesson to review each one.

• The distance formula given by $d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$d=√(x2​−x1​)2+(y2​−y1​)2
• The slope formula $m=\frac{y_2-y_1}{x_2-x_1}$m=y2​−y1​x2​−x1​​
• The perpendicular property $m_1*m_2=-1$m1​*m2​=−1
• The parallel property $m_1=m_2$m1​=m2​
• The midpoint formula is given by $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$(x1​+x2​2​,y1​+y2​2​)

In the lesson, we will focus on triangles and quadrilaterals. You should be familiar with the properties of side lengths and angles of the following polygons.

 

Worked examples

Checking a parallelogram

Show that the quadrilateral with vertices given by $P\left(2,3\right),Q\left(3,6\right),R\left(6,8\right),S\left(5,5\right)$P(2,3),Q(3,6),R(6,8),S(5,5) is a parallelogram.

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

Think: If opposite sides are parallel then they must have the same slope. We need $\overline{PQ}\parallel\overline{RS}$PQ∥RS and $\overline{QR}\parallel\overline{PS}$QR∥PS

Do: We can check the slopes of $\overline{PQ}$PQ and $\overline{RS}$RS and the slopes of the line segments $\overline{QR}$QR and $\overline{PS}$PS.

Reflect: Both pairs of opposite sides are parallel. Hence, the quadrilateral is a parallelogram.

 

Checking a right triangle

Prove that $A\left(-4,-4\right),B\left(-1,-1\right),C\left(1,-3\right)$A(−4,−4),B(−1,−1),C(1,−3) are the vertices of a right triangle.

Think: To be a right triangle, we need to have one pair of perpendicular sides. The easiest way to do this is to check for the perpendicular property $m_1m_2=-1$m1​m2​=−1 given above for $m_{\overline{AB}}$mAB​, $m_{\overline{AB}}$mAB​, and $m_{\overline{AB}}$mAB​.

Do: The three slopes are determined as

$m_{\overline{AB}}=\frac{-1-\left(-4\right)}{-1-\left(-4\right)}=1$mAB​=−1−(−4)−1−(−4)​=1

$m_{\overline{BC}}=\frac{-3-\left(-1\right)}{1-\left(-1\right)}=-1$mBC​=−3−(−1)1−(−1)​=−1

$m_{\overline{AC}}=\frac{-3-\left(-4\right)}{1-\left(-4\right)}=\frac{1}{5}$mAC​=−3−(−4)1−(−4)​=15​

Reflect: Note that $m_{\overline{AB}}\times m_{\overline{BC}}=-1$mAB​×mBC​=−1 and so $\overline{BC}$BC is perpendicular to $\overline{AB}$AB, and the triangle is right-angled.  

 

Checking a rhombus

Prove that the quadrilateral with vertices $A\left(4,9\right),B\left(5,13\right),C\left(9,14\right),D\left(8,10\right)$A(4,9),B(5,13),C(9,14),D(8,10) is a rhombus.

Think: A rhombus must have all sides of equal length and opposite sides must be parallel. It does not need to be a square, so the adjacent sides do not need to be perpendicular.

Do: Let's look at distances first, to ensure that they are all the same. We find:

$\overline{AB}=\sqrt{1^2+4^2}=\sqrt{17}$AB=√12+42=√17

$\overline{BC}=\sqrt{4^2+1^2}=\sqrt{17}$BC=√42+12=√17

$\overline{CD}=\sqrt{1^2+4^2}=\sqrt{17}$CD=√12+42=√17

$\overline{DA}=\sqrt{4^2+1^2}=\sqrt{17}$DA=√42+12=√17

Next, let's calculate slope to ensure that opposite sides are parallel,

$m_{\overline{AB}}=\frac{4}{1}=4$mAB​=41​=4

$m_{\overline{CD}}=\frac{-4}{-1}=4$mCD​=−4−1​=4

so $m_{\overline{AB}}\parallel m_{\overline{CD}}$mAB​∥mCD​

$m_{\overline{BC}}=\frac{1}{4}$mBC​=14​

$m_{\overline{DA}}=\frac{1}{4}$mDA​=14​

so $m_{\overline{BC}}\parallel m_{\overline{DA}}$mBC​∥mDA​

Reflect: The opposite side pairs are parallel and all sides are the same length, so this is indeed a rhombus.

 

Practice questions

Question 1

Question 2

Question 3