Solving Geometry Problems on Coordinate Axes

Calculating the Distance Between Two Points

The distance between two points on a two dimensional coordinate plane can be determined by applying Pythagoras theorem. If we consider two points, $A(a, b)$ and $C(c, d)$, we can visualize a right-angled triangle by drawing a vertical line through $C$ and a horizontal line through $A$. These lines meet at a point $B$.

In this triangle, the horizontal distance $AB$ is the difference between the x-coordinates, while the vertical distance $BC$ is the difference between the y-coordinates. According to Pythagoras theorem, the square of the hypotenuse $AC$ is equal to the sum of the squares of the other two sides:

$AC^2 = AB^2 + BC^2$

Substituting the coordinate differences into this formula gives:

$AC^2 = (c - a)^2 + (d - b)^2$

This can also be written as:

(distance between two points)$^2$ = (difference in y-coordinates)$^2$ + (difference in x-coordinates)$^2$

Worked Example: Finding Distance

Point $A$ is $(2, -3)$ and point $B$ is $(-4, 5)$. To find the distance $AB$:

1. Calculate the squared difference in y-coordinates: $(5 - (-3))^2 = 8^2 = 64$.
2. Calculate the squared difference in x-coordinates: $(-4 - 2)^2 = (-6)^2 = 36$.
3. Sum these values: $64 + 36 = 100$.
4. Take the square root: $AB = \sqrt{100} = 10$.

Finding Midpoints

The midpoint of a line segment is the point that lies exactly halfway between two given coordinates. It represents the average of the positions. For points $A(a, b)$ and $C(c, d)$, the midpoint is calculated as:

$\left(\frac{a + c}{2}, \frac{b + d}{2}\right)$

Worked Example: Finding a Midpoint

Point $A$ is $(2, -3)$ and point $B$ is $(-4, 5)$. If $C$ is the midpoint of $AB$:

1. The x-coordinate of $C$ is $\frac{2 + (-4)}{2} = -\frac{2}{2} = -1$.
2. The y-coordinate of $C$ is $\frac{-3 + 5}{2} = \frac{2}{2} = 1$.
3. The coordinates of $C$ are $(-1, 1)$.

Problem Solving with Coordinates

More complex geometric problems can be solved by translating shape properties into coordinate relationships. Success in these problems often depends on a structured approach.

1. Always draw a diagram. It does not need to be perfectly accurate, but it must show the relative positions of the points correctly.
2. Ensure the vertex labels (such as $A, B, C, D$) follow a consistent order, either clockwise or anticlockwise, around the shape.
3. Identify geometric traits like parallel lines (which share a gradient) or perpendicular lines.

Worked Example: Trapezium Problem

$ABCD$ is a trapezium where $AB$ is parallel to $DC$. Point $A$ is $(-2, 4)$, $B$ is $(5, 4)$, and $D$ is $(-2, -3)$. We are given that angle $BCD$ is an acute angle and the length of $BC$ is $12$. We must find the length of $DC$.

When sketching the diagram, several key observations appear:

• $D$ is on the same vertical line as $A$ (both have x-coordinate $-2$).
• $B$ is on the same horizontal line as $A$ (both have y-coordinate $4$).
• This means the angle at $A$ is 90 degrees. Because $AB$ is parallel to $DC$, the angle at $D$ is also 90 degrees.
• By drawing a vertical line $BE$ parallel to $AD$, we create a rectangle $ABED$ and a right-angled triangle $BEC$.

From these observations, we can calculate lengths:

1. $BE = AD = 4 - (-3) = 7$.
2. $DE = AB = 5 - (-2) = 7$.
3. Since angle $BCD$ is acute, $C$ must be further right than $E$, meaning $DC$ is longer than $AB$.

Now, apply Pythagoras theorem to the right-angled triangle $BEC$:

$EC^2 = BC^2 - BE^2 = 12^2 - 7^2 = 144 - 49 = 95$. $EC = \sqrt{95}$.

Finally, the total length of $DC$ is the sum of $DE$ and $EC$:

$DC = 7 + \sqrt{95}$.