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:
Here are some examples:
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 gradients of the line segments $PQ$PQ and $RS$RS and the gradients 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.
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 $m_1m_2=-1$m1m2=−1 given above.
The three gradients 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.
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 gradients,
$m_{AB}=\frac{4}{1}=4,m_{BC}=\frac{1}{4}$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 gradients 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.
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.
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.
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 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.
The equation of the line through $O$O and $Q$Q (passing through the origin) is given by:
$y$y | $=$= | $\frac{c}{b}x$cbx $(1)$(1) |
This is because the gradient 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 gradient 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:
$\frac{c}{b}x$cbx | $=$= | $\left(\frac{c}{b-3a}\right)\left(x-a\right)$(cb−3a)(x−a) |
$\frac{1}{b}x$1bx | $=$= | $\left(\frac{1}{b-3a}\right)\left(x-a\right)$(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=cbx=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.
To do this we simply show that the gradient 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 gradient formula, we have for the gradient 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 gradient 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.
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.
Consider the triangle shown below:
Determine the gradient of the line segment $AB$AB.
Similarly, determine the gradient of side $AC$AC:
Next determine the length of the side $AB$AB.
Now determine the length of the side $AC$AC.
Hence state the type of triangle that has been graphed. Choose the most correct answer:
An equilateral triangle.
An acute isosceles triangle.
An isosceles right-angled triangle.
A scalene right-angled triangle.
An equilateral triangle.
An acute isosceles triangle.
An isosceles right-angled triangle.
A scalene right-angled triangle.
Consider the quadrilateral shown below:
First determine the gradient of the line segment $AB$AB.
Next determine the gradient of side $CD$CD.
Similarly, determine the gradients of the other two sides:
Gradient of $AD$AD $=$= $\editable{}$
Gradient of $BC$BC $=$= $\editable{}$
Determine the length of the side $AB$AB.
Now determine the length of the side $AD$AD.
Hence state the type of quadrilateral that has been graphed. Choose the most correct answer.
Rectangle
Rhombus
Square
Trapezium
Rectangle
Rhombus
Square
Trapezium
Given Line P: $y=-6x-4$y=−6x−4, Line Q: $y=\frac{x}{6}+6$y=x6+6, Line R: $y=-6x-1$y=−6x−1 and Line S: $y=\frac{x}{6}+1$y=x6+1.
Complete the following:
$m$m_{P }= $\editable{}$
$m$m_{Q }= $\editable{}$
$m$m_{P} x $m$m_{Q }= $\editable{}$
Complete the following:
$m$m_{Q }= $\editable{}$
$m$m_{R }= $\editable{}$
$m$m_{Q} x $m$m_{R }= $\editable{}$
Complete the following:
$m$m_{R }= $\editable{}$
$m$m_{S }= $\editable{}$
$m$m_{R} x $m$m_{S }= $\editable{}$
Complete the following:
$m$m_{S }= $\editable{}$
$m$m_{P }= $\editable{}$
$m$m_{S} x $m$m_{P }= $\editable{}$
What type of quadrilateral is formed by lines: P, Q, R, and S?
Trapezoid
Rectangle
Rhombus
Parallelogram
Trapezoid
Rectangle
Rhombus
Parallelogram
Apply co-ordinate geometry techniques to points and lines
Apply co-ordinate geometry methods in solving problems