Applications in 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 results:
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 midpoint formula is given by $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$M=(x1+x22,y1+y22)
The gradient formula $m=\frac{y_2-y_1}{x_2-x_1}$m=y2−y1x2−x1
The parallel property, if line $1$1: $y=m_1x+c_1$y=m1x+c1 is parallel to line $2$2: $y=m_2x+c_2$y=m2x+c2 then $m_1=m_2$m1=m2
The perpendicular property, if line $1$1: $y=m_1x+c_1$y=m1x+c1 is perpendicular to line $2$2: $y=m_2x+c_2$y=m2x+c2 then $m_2=\frac{-1}{m_1}$m2=−1m1
Practice questions
Question 1
The endpoints of the diameter of a circle are $\left(4,9\right)$(4,9) and $\left(0,3\right)$(0,3).
Find the coordinates of the centre.
centre $=$= $($( $\editable{}$, $\editable{}$ $)$)
Find the radius of the circle, leaving your answer to two decimal places.
Question 2
Consider the Points $A$A $\left(12,6\right)$(12,6) and $B$B $\left(7,4\right)$(7,4).
Find the exact length of $AB$AB (that is, leaving your answer in surd form).
Given that $M$M is the midpoint of $AB$AB, find the exact length of $AM$AM (that is, leaving your answer in surd form).
Question 3
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 exact length of the side $AB$AB.
Now determine the exact length of the side $AC$AC.
Hence state the type of triangle that has been graphed. Choose the most correct answer:
An equilateral triangle.
AAn acute isosceles triangle.
BAn isosceles right-angled triangle.
CA scalene right-angled triangle.
D
Parallel and perpendicular lines
Two lines that are parallel have the same gradient.
Thus two lines given by $y=m_1x+c_1$y=m1x+c1 and $y=m_2x+c_2$y=m2x+c2 will be parallel if and only if $m_1=m_2$m1=m2.
Two lines are perpendicular if the product of their respective gradients equals $-1$−1. Another way to say this is that the gradient of one line is the negative reciprocal of the other.
Thus two lines given by $y=m_1x+c_1$y=m1x+c1 and $y=m_2x+c_2$y=m2x+c2 will be perpendicular if and only if $m_1=-\frac{1}{m_2}$m1=−1m2.
We can explain these features with the following diagram:
|
Parallel property
$m_1$m1$=$=$m_2$m2$=$=$\frac{p}{q}$pq |
Perpendicular property
$m_1=\frac{a}{b}$m1=ab $m_2=\frac{-b}{a}$m2=−ba |
Practice questions
Question 4
Find the equation of the line $L_1$L1 that is parallel to the line $y=-\frac{2x}{7}+1$y=−2x7+1 and goes through the point $\left(0,-10\right)$(0,−10). Give your answer in the form $y=mx+b$y=mx+b.
Question 5
Consider the points $A$A$\left(5,6\right)$(5,6) and $B$B$\left(-13,-22\right)$(−13,−22).
Find the gradient of interval $AB$AB.
Find the midpoint of $AB$AB.
Midpoint $=$= $\left(\editable{},\editable{}\right)$(,)
Find the equation of the perpendicular bisector of $AB$AB. That is, find the line perpendicular to $AB$AB and passing through its midpoint.
Express your answer in general form, $ax+by+c=0$ax+by+c=0.