Geometrical problems in coordinate geometry

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

There is a range of established results that become useful in this endeavour. These include the following six results:

1. The distance formula given by $d=\sqrt{\left(x-x_1\right)^2+\left(y-y_1\right)^2}$d=√(x−x1​)2+(y−y1​)2
2. The slope formula $m=\frac{y_2-y_1}{x_2-x_1}$m=y2​−y1​x2​−x1​​
3. The perpendicular property $m_1m_2=-1$m1​m2​=−1 and the parallel property $m_1=m_2$m1​=m2​
4. 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​)
5. The point/slope form for the equation of a line $y-y_1=m\left(x-x_1\right)$y−y1​=m(x−x1​)
6. The perpendicular distance formula $d=\frac{\left|ax_1+by_1+c\right|}{\sqrt{a^2+b^2}}$d=|ax1​+by1​+c|√a2+b2​.

 

Here are some examples:

Example 1:

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 closed four sided figure with opposite sides parallel. We can check the slopes of the line segments $PQ$PQ and $RS$RS and the slopes of the line segments $QR$QR and $PS$PS.

$m_{PQ}=\frac{6-3}{3-2}=3,m_{RS}=\frac{5-8}{5-6}=3$mPQ​=6−33−2​=3,mRS​=5−85−6​=3

$m_{QR}=\frac{8-6}{6-3}=\frac{2}{3},m_{PS}=\frac{5-3}{5-2}=\frac{2}{3}$mQR​=8−66−3​=23​,mPS​=5−35−2​=23​.

Hence, the quadrilateral is a parallelogram.

 

Example 2:

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-angled triangle.

The easiest way to do this is to use the perpendicular property $m_1m_2=-1$m1​m2​=−1 given above.

The three slopes are determined as $m_{AB}=\frac{-1-\left(-4\right)}{-1-\left(-4\right)}=1$mAB​=−1−(−4)−1−(−4)​=1, $m_{BC}=\frac{-3-\left(-1\right)}{1-\left(-1\right)}=-1$mBC​=−3−(−1)1−(−1)​=−1 and $m_{AC}=\frac{-3-\left(-4\right)}{1-\left(-4\right)}=\frac{1}{5}$mAC​=−3−(−4)1−(−4)​=15​. 

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

 

Example 3:

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.

If we look at distances first, we find:

$d_{AB}=\sqrt{1^2+4^2}=\sqrt{17},d_{BC}=\sqrt{4^2+1^2}=\sqrt{17}$dAB​=√12+42=√17,dBC​=√42+12=√17

$d_{CD}=\sqrt{1^2+4^2}=\sqrt{17},d_{DA}=\sqrt{4^2+1^2}=\sqrt{17}$dCD​=√12+42=√17,dDA​=√42+12=√17

Also, evaluating slopes,

$m_{AB}=\frac{4}{1}=4,m_{BC}=\frac{1}{4}$mAB​=41​=4,mBC​=14​

$m_{CD}=\frac{-4}{-1}=4,m_{DA}=\frac{1}{4}$mCD​=−4−1​=4,mDA​=14​

The product of any two of the slopes is never $-1$−1, and so the figure is not a square. In fact it is a parallelogram with equal sides, making it a rhombus.

 

An extension example

Prove that the medians of any triangle meet at an interior common point $G$G.

This is a challenging proof, but a proof worth going through, because it makes use of many of the coordinate geometry results listed above.  

Describing the problem

A median is a line segment drawn from a vertex to the midpoint of the opposite side. Thus there are three medians of a triangle, and it can be proved using a number of methods that the three medians intersect each other at a common point $G$G called the centroid (also called the centre of gravity).

One of these proofs uses coordinate geometry, and will be given below.

The diagram

While it is a completely general proof, we can reduce the algebra involved by setting the triangle up so that one edge lies along the $x$x axis in such a way as to put the midpoint of that edge at the origin. 

We then can label the vertices $P\left(-a,0\right),Q\left(b,c\right)$P(−a,0),Q(b,c) and $R\left(a,0\right)$R(a,0), as shown in the diagram:

The midpoints $D$D, $E$E and $O$O and the three medians have been added to the diagram as well, although at this stage, we haven't proved that they are concurrent at $G$G.

The plan of attack

The first step in the proof is to find the coordinates of $G$G using two of the medians (the ones shown in black) 

The second step is to show that the third median (in blue) also passes through $G$G.

Finding G

The equation of the line through $O$O and $Q$Q (passing through the origin) is given by:

$y$y	$=$=	$\frac{c}{b}x$cb​x    $(1)$(1)

This is because the slope is given by $m_1=\frac{c}{b}$m1​=cb​ and the $y$y intercept is $0$0.

The coordinates of $D$D, using the mid-point formula, become $\left(\frac{b-a}{2},\frac{c}{2}\right)$(b−a2​,c2​). 

The slope of the line through $D$D and $R$R is then $m_2=\left(\frac{\frac{c}{2}-0}{\frac{b-a}{2}-a}\right)$m2​=(c2​−0b−a2​−a​).

By multiplying the numerator and denominator by $2$2, this simplifies to $\left(\frac{c}{b-3a}\right)$(cb−3a​).

Then, using the point $R\left(a,0\right)$R(a,0) and $m_2$m2​, the equation of the line through $D$D and $R$R becomes:

$y$y	$=$=	$\left(\frac{c}{b-3a}\right)\left(x-a\right)$(cb−3a​)(x−a)     $(2)$(2)

Finally, we need to solve simultaneously equations $(1)$(1) and $(2)$(2) to find the coordinates of $G$G.

Thus we set the two right hand sides equal to each other and solve for $x$x:

$\frac{c}{b}x$cb​x	$=$=	$\left(\frac{c}{b-3a}\right)\left(x-a\right)$(cb−3a​)(x−a)
$\frac{1}{b}x$1b​x	$=$=	$\left(\frac{1}{b-3a}\right)\left(x-a\right)$(1b−3a​)(x−a)
$bx-3ax$bx−3ax	$=$=	$bx-ab$bx−ab
$-3ax$−3ax	$=$=	$-ab$−ab
$\therefore$∴     $x$x	$=$=	$\frac{b}{3}$b3​

With $x=\frac{b}{3}$x=b3​, substituting back into $y$y shows $y=\frac{c}{b}x=\frac{c}{b}\times\frac{b}{3}=\frac{c}{3}$y=cb​x=cb​×b3​=c3​.

So the coordinates of $G$G are $\left(\frac{b}{3},\frac{c}{3}\right)$(b3​,c3​), but we still have a little further to go. We still haven't shown that the third median $PE$PE intersects the other medians at $G$G. 

Proving concurrency

To do this we simply show that the slope of $PE$PE is the same as $PG$PG.

The coordinates of $E$E, using the midpoint formula is $\left(\frac{b+a}{2},\frac{c}{2}\right)$(b+a2​,c2​).

Using our slope formula, we have for the slope of $PE$PE:

$m_{PE}$mPE​	$=$=	$\frac{\frac{c}{2}-0}{\frac{b+a}{2}-\left(-a\right)}$c2​−0b+a2​−(−a)​
	$=$=	$\frac{c}{b+a+2a}$cb+a+2a​
	$=$=	$\frac{c}{b+3a}$cb+3a​

and the slope of $PG$PG is:

$m_{PG}$mPG​	$=$=	$\frac{\frac{c}{3}-0}{\frac{b}{3}-\left(-a\right)}$c3​−0b3​−(−a)​
	$=$=	$\frac{c}{b+3a}$cb+3a​

This has to imply that all of the medians are concurrent.

Centre of Gravity

Because our proof is a general one, we learn an amazing fact about the centroid.

If we consider the point $S$S whose coordinates are the average the $x$x and $y$y parts of the coordinates of the vertices $P\left(-a,0\right),Q\left(b,c\right)$P(−a,0),Q(b,c) and $R\left(a,0\right)$R(a,0), we find that:

$S=\left(\frac{-a+b+a}{3},\frac{0+c+0}{3}\right)=\left(\frac{b}{3},\frac{c}{3}\right)=G$S=(−a+b+a3​,0+c+03​)=(b3​,c3​)=G

In other words, the coordinates of the centroid are found by averaging the coordinates of the vertices!

In fact the centroid is the balance point of the triangle, and that's why it is also called the centre of gravity.   

 

Worked Examples

Question 1

Question 2

Question 3