6.3 Geometry  (Page 4/7)

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:

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 $(x_1;y_1)$ and $B$ is any other point with co-ordinates $(x_2;y_2)$ .

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:

However,

Therefore,

Therefore, for any two points, $(x_1;y_1)$ and $(x_2;y_2)$ , the formula is:

Distance= $\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

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 $(x_1;y_1)$ and the co-ordinates of point $Q$ be $(x_2;y_2)$ . Then the distance is:

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 $(x_1;y_1)$ and a point $B$ with co-ordinates $(x_2;y_2)$ .

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:

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 $(2;1)$ and point $Q$ with co-ordinates $(-2;-2)$ .

The co-ordinates of the midpoint of any line between any two points $A$ and $B$ with co-ordinates $(x_1;y_1)$ and $(x_2;y_2)$ , is generally calculated as follows. Let the midpoint of $AB$ be at point $S$ with co-ordinates $(X;Y)$ . The aim is to calculate $X$ and $Y$ in terms of $(x_1;y_1)$ and $(x_2;y_2)$ .

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

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