Distance and Midpoints (VCE SSCE Mathematical Methods): Revision Notes

Distance and Midpoints

In this note, we explore two fundamental techniques in coordinate geometry: finding the midpoint of a line segment and calculating the distance between two points. These skills form the foundation for many coordinate geometry problems you'll encounter.

Midpoint of a line segment

The midpoint of a line segment is the point that lies exactly halfway between the two endpoints. Finding this point involves averaging the coordinates of the endpoints.

Line segments parallel to an axis

When a line segment is parallel to one of the axes (either horizontal or vertical), finding the midpoint becomes straightforward.

Vertical line segments

Consider a vertical line segment from $A(2, 3)$ down to $B(2, -4)$. Since the line is vertical, the $x$-coordinate remains constant at $2$. The midpoint $P$ therefore has coordinates $(2, -\frac{1}{2})$.

Notice that $-\frac{1}{2}$ is exactly the average of the $y$-coordinates:

$$\frac{3 + (-4)}{2} = -\frac{1}{2}$$

Horizontal line segments

Similarly, for a horizontal line segment from $C(-1, 2)$ to $D(3, 2)$, the $y$-coordinate stays constant at $2$. The midpoint $P$ has coordinates $(1, 2)$.

Here, $1$ is the average of the $x$-coordinates:

$$\frac{-1 + 3}{2} = 1$$

Line segments not parallel to axes

For a line segment that isn't parallel to either axis, we need a more general approach. Let's find the midpoint $P(x, y)$ of a line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$.

We can construct a clever proof by drawing two right triangles. First, draw horizontal and vertical lines from point $A$ to create point $C$, and similarly from point $P$ to create point $D$. This creates two triangles: $APC$ and $PBD$.

These two triangles are identical in size and shape (congruent). This means their corresponding sides are equal:

$$AC = PD \text{ and } PC = BD$$

In terms of coordinates, this translates to:

$$x - x_1 = x_2 - x$$ $$y - y_1 = y_2 - y$$

Rearranging the first equation:

$$2x = x_1 + x_2$$ $$x = \frac{x_1 + x_2}{2}$$

Rearranging the second equation:

$$2y = y_1 + y_2$$ $$y = \frac{y_1 + y_2}{2}$$

This gives us the general midpoint formula.

Worked example

Distance between two points

To find the distance between two points on a coordinate plane, we apply Pythagoras' theorem. This theorem states that in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

Deriving the distance formula

Consider two points $A(x_1, y_1)$ and $B(x_2, y_2)$. We can form a right-angled triangle by drawing a horizontal line from $A$ and a vertical line from $B$ to meet at point $C$.

The horizontal distance (base of triangle) is $x_2 - x_1$

The vertical distance (height of triangle) is $y_2 - y_1$

The distance $AB$ is the hypotenuse of this right triangle.

Applying Pythagoras' theorem:

$$AB^2 = AC^2 + BC^2$$ $$AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$

Taking the square root of both sides:

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

Worked example

Remember!