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VCE 11 Methods 2023

1.08 Coordinate geometry

Lesson

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. 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=(x2x1)2+(y2y1)2

The midpoint formula is given by $M=$M= $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$(x1+x22,y1+y22)

The gradient formula $m=\frac{y_2-y_1}{x_2-x_1}$m=y2y1x2x1

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).

  1. Find the coordinates of the centre.

    centre $=$= $($( $\editable{}$, $\editable{}$ $)$)

  2. 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).

  1. Find the exact length of $AB$AB (that is, leaving your answer in surd form).

  2. 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:

Loading Graph...

  1. Determine the gradient of the line segment $AB$AB.

  2. Similarly, determine the gradient of side $AC$AC:

  3. Next determine the exact length of the side $AB$AB.

  4. Now determine the exact length of the side $AC$AC.

  5. Hence state the type of triangle that has been graphed. Choose the most correct answer:

    An equilateral triangle.

    A

    An acute isosceles triangle.

    B

    An isosceles right-angled triangle.

    C

    A 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.

These features can be explained 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).

  1. Find the gradient of interval $AB$AB.

  2. Find the midpoint of $AB$AB.

    Midpoint $=$=  $\left(\editable{},\editable{}\right)$(,)

  3. 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.

 

Outcomes

U1.AoS2.8

solution of a set of simultaneous linear equations (geometric interpretation only required for two variables) and equations of the form f(x) = g(x) numerically, graphically and algebraically.

U1.AoS3.1

average and instantaneous rates of change in a variety of practical contexts and informal treatment of instantaneous rate of change as a limiting case of the average rate of change

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