# 7.2 Analytical geometry

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## Introduction

Analytical geometry, also called co-ordinate geometry and earlier referred to as Cartesian geometry, is the study of geometry using the principles of algebra, and the Cartesian co-ordinate system. It is concerned with defining geometrical shapes in a numerical way, and extracting numerical information from that representation. Some consider that the introduction of analytic geometry was the beginning of modern mathematics.

## Distance between two points

One of the simplest things that can be done with analytical geometry is to calculate the distance between two points. Distance is a number that describes how far apart two point are. For example, point $P$ has co-ordinates $\left(2,1\right)$ and point $Q$ has co-ordinates $\left(-2,-2\right)$ . How far apart are points $P$ and $Q$ ? In the figure, this means how long is the dashed line?

In the figure, it can be seen that the length of the line $PR$ is 3 units and the length of the line $QR$ is four units. However, the $▵PQR$ , has a right angle at $R$ . Therefore, the length of the side $PQ$ can be obtained by using the Theorem of Pythagoras:

$\begin{array}{ccc}\hfill P{Q}^{2}& =& P{R}^{2}+Q{R}^{2}\hfill \\ \hfill \therefore P{Q}^{2}& =& {3}^{2}+{4}^{2}\hfill \\ \hfill \therefore PQ& =& \sqrt{{3}^{2}+{4}^{2}}=5\hfill \end{array}$

The length of $PQ$ is the distance between the points $P$ and $Q$ .

In order to generalise the idea, assume $A$ is any point with co-ordinates $\left({x}_{1};{y}_{1}\right)$ and $B$ is any other point with co-ordinates $\left({x}_{2};{y}_{2}\right)$ .

The formula for calculating the distance between two points is derived as follows. The distance between the points $A$ and $B$ is the length of the line $AB$ . According to the Theorem of Pythagoras, the length of $AB$ is given by:

$AB=\sqrt{A{C}^{2}+B{C}^{2}}$

However,

$\begin{array}{c}\hfill BC={y}_{2}-{y}_{1}\\ \hfill AC={x}_{2}-{x}_{1}\end{array}$

Therefore,

$\begin{array}{ccc}\hfill AB& =& \sqrt{A{C}^{2}+B{C}^{2}}\hfill \\ & =& \sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}\hfill \end{array}$

Therefore, for any two points, $\left({x}_{1};{y}_{1}\right)$ and $\left({x}_{2};{y}_{2}\right)$ , the formula is:

Distance= $\sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}$

Using the formula, distance between the points $P$ and $Q$ with co-ordinates (2;1) and (-2;-2) is then found as follows. Let the co-ordinates of point $P$ be $\left({x}_{1};{y}_{1}\right)$ and the co-ordinates of point $Q$ be $\left({x}_{2};{y}_{2}\right)$ . Then the distance is:

$\begin{array}{ccc}\hfill \mathrm{Distance}& =& \sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}\hfill \\ & =& \sqrt{{\left(2-\left(-2\right)\right)}^{2}+{\left(1-\left(-2\right)\right)}^{2}}\hfill \\ & =& \sqrt{{\left(2+2\right)}^{2}+{\left(1+2\right)}^{2}}\hfill \\ & =& \sqrt{16+9}\hfill \\ & =& \sqrt{25}\hfill \\ & =& 5\hfill \end{array}$

The following video provides a summary of the distance formula.

## Calculation of the gradient of a line

The gradient of a line describes how steep the line is. In the figure, line $PT$ is the steepest. Line $PS$ is less steep than $PT$ but is steeper than $PR$ , and line $PR$ is steeper than $PQ$ .

The gradient of a line is defined as the ratio of the vertical distance to the horizontal distance. This can be understood by looking at the line as the hypotenuse of a right-angled triangle. Then the gradient is the ratio of the length of the vertical side of the triangle to the horizontal side of the triangle. Consider a line between a point $A$ with co-ordinates $\left({x}_{1};{y}_{1}\right)$ and a point $B$ with co-ordinates $\left({x}_{2};{y}_{2}\right)$ .

Gradient $=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

We can use the gradient of a line to determine if two lines are parallel or perpendicular. If the lines are parallel ( [link] a) then they will have the same gradient, i.e. m AB $=$ m CD . If the lines are perpendicular ( [link] b) than we have: $-\frac{1}{{m}_{\mathrm{AB}}}={m}_{\mathrm{CD}}$

For example the gradient of the line between the points $P$ and $Q$ , with co-ordinates (2;1) and (-2;-2) ( [link] ) is:

$\begin{array}{ccc}\hfill \mathrm{Gradient}& =& \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\hfill \\ & =& \frac{-2-1}{-2-2}\hfill \\ & =& \frac{-3}{-4}\hfill \\ & =& \frac{3}{4}\hfill \end{array}$

The following video provides a summary of the gradient of a line.

## Midpoint of a line

Sometimes, knowing the co-ordinates of the middle point or midpoint of a line is useful. For example, what is the midpoint of the line between point $P$ with co-ordinates $\left(2;1\right)$ and point $Q$ with co-ordinates $\left(-2;-2\right)$ .

The co-ordinates of the midpoint of any line between any two points $A$ and $B$ with co-ordinates $\left({x}_{1};{y}_{1}\right)$ and $\left({x}_{2};{y}_{2}\right)$ , is generally calculated as follows. Let the midpoint of $AB$ be at point $S$ with co-ordinates $\left(X;Y\right)$ . The aim is to calculate $X$ and $Y$ in terms of $\left({x}_{1};{y}_{1}\right)$ and $\left({x}_{2};{y}_{2}\right)$ .

$\begin{array}{ccc}\hfill X& =& \frac{{x}_{1}+{x}_{2}}{2}\hfill \\ \hfill Y& =& \frac{{y}_{1}+{y}_{2}}{2}\hfill \\ \hfill \therefore & & S\left(\frac{{x}_{1}+{x}_{2}}{2},;,\frac{{y}_{1}+{y}_{2}}{2}\right)\hfill \end{array}$

Then the co-ordinates of the midpoint ( $S$ ) of the line between point $P$ with co-ordinates $\left(2;1\right)$ and point $Q$ with co-ordinates $\left(-2;-2\right)$ is:

$\begin{array}{ccc}\hfill X& =& \frac{{x}_{1}+{x}_{2}}{2}\hfill \\ & =& \frac{-2+2}{2}\hfill \\ & =& 0\hfill \\ \hfill Y& =& \frac{{y}_{1}+{y}_{2}}{2}\hfill \\ & =& \frac{-2+1}{2}\hfill \\ & =& -\frac{1}{2}\hfill \\ \hfill \therefore S\phantom{\rule{4pt}{0ex}}\mathrm{is}\phantom{\rule{4pt}{0ex}}\mathrm{at}\phantom{\rule{4pt}{0ex}}\left(0;-\frac{1}{2}\right)\end{array}$

It can be confirmed that the distance from each end point to the midpoint is equal. The co-ordinate of the midpoint $S$ is $\left(0;-0,5\right)$ .

$\begin{array}{ccc}\hfill PS& =& \sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}\hfill \\ & =& \sqrt{{\left(0-2\right)}^{2}+{\left(-0.5-1\right)}^{2}}\hfill \\ & =& \sqrt{{\left(-2\right)}^{2}+{\left(-1.5\right)}^{2}}\hfill \\ & =& \sqrt{4+2.25}\hfill \\ & =& \sqrt{6.25}\hfill \end{array}$

and

$\begin{array}{ccc}\hfill QS& =& \sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}\hfill \\ & =& \sqrt{{\left(0-\left(-2\right)\right)}^{2}+{\left(-0.5-\left(-2\right)\right)}^{2}}\hfill \\ & =& \sqrt{{\left(0+2\right)\right)}^{2}{+\left(-0.5+2\right)\right)}^{2}}\hfill \\ & =& \sqrt{{\left(2\right)\right)}^{2}{+\left(-1.5\right)\right)}^{2}}\hfill \\ & =& \sqrt{4+2.25}\hfill \\ & =& \sqrt{6.25}\hfill \end{array}$

It can be seen that $PS=QS$ as expected.

The following video provides a summary of the midpoint of a line.

## Co-ordinate geometry

1. In the diagram given the vertices of a quadrilateral are F(2;0), G(1;5), H(3;7) and I(7;2).
1. What are the lengths of the opposite sides of FGHI?
2. Are the opposite sides of FGHI parallel?
3. Do the diagonals of FGHI bisect each other?
2. A quadrialteral ABCD with vertices A(3;2), B(1;7), C(4;5) and D(1;3) is given.
2. Find the lengths of the sides of the quadrilateral.
3. ABCD is a quadrilateral with verticies A(0;3), B(4;3), C(5;-1) and D(-1;-1).
1. Show that:
2. AB $\parallel$ DC
2. What name would you give to ABCD?
3. Show that the diagonals AC and BD do not bisect each other.
4. P, Q, R and S are the points (-2;0), (2;3), (5;3), (-3;-3) respectively.
1. Show that:
1. SR = 2PQ
2. SR $\parallel$ PQ
2. Calculate:
1. PS
2. QR
5. EFGH is a parallelogram with verticies E(-1;2), F(-2;-1) and G(2;0). Find the co-ordinates of H by using the fact that the diagonals of a parallelogram bisect each other.

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