Coordinate Geometry | Sec 3 & 4 A Maths Study Guide

Secondary 3 & 4 Additional Mathematics · Study guide

Connect geometric conditions to coordinate equations, then check lengths, vertices and areas.

Concept → worked examples → practice20 min read · practice linked below

Turn a geometrical condition into an equation

This shared Secondary 3 and 4 guide covers straight lines, lengths, division of a segment and areas. A diagram helps you see the relationships; coordinates let you justify them. Identify the condition first: parallel, perpendicular, equidistant, midpoint or specified area. Then choose the equation that expresses that condition.

Tools for points $A(x_1,y_1)$ and $B(x_2,y_2)$
Quantity	Formula and reason
Length	$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , from Pythagoras applied to the horizontal and vertical changes.
Midpoint	$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ , the average of each coordinate.
Gradient	$m=\frac{y_2-y_1}{x_2-x_1}$ , provided $x_2\ne x_1$ . Use the same direction of subtraction in numerator and denominator.
Line through $(p,q)$	$y-q=m(x-p)$ for a finite gradient. A vertical line has equation $x=p$ .

Distinct nonvertical lines are parallel when their gradients are equal. Perpendicular lines with finite gradients satisfy $m_1m_2=-1$ . A horizontal line and a vertical line are also perpendicular; handle them directly because a vertical gradient is undefined. Three distinct points are collinear when the changes between them have the same direction, including the vertical case.

Example 1: length, midpoint and a line

Find the length and midpoint of $A(-2,1)$ , $B(4,9)$ , and the equation of $AB$ .

The changes are $6$ horizontally and $8$ vertically, so $AB=\sqrt{6^2+8^2}=10$ .
The midpoint is $M=(\frac{-2+4}{2},\frac{1+9}{2})=(1,5)$ .
The gradient is $8/6=4/3$ . Using point $A$ , $y-1=\frac43(x+2)$ , hence $4x-3y+11=0$ .

Check both endpoints: $4(-2)-3(1)+11=0$ and $4(4)-3(9)+11=0$ . The midpoint also satisfies the line equation.

Perpendicular bisectors and unknown points

The perpendicular bisector of $AB$ passes through its midpoint and is perpendicular to $AB$ . Every point on it is equidistant from $A$ and $B$ : the two right triangles formed with the midpoint have equal legs. Conversely, expanding $PA^2=PB^2$ cancels the squared coordinates and produces this same line.

Example 2: equidistant from two points

Find the point $P$ on $y=x+2$ that is equidistant from the points $A(-2,1)$ and $B(4,9)$ .

From Example 1, the midpoint is $(1,5)$ and $m_{AB}=4/3$ . The perpendicular gradient is $-3/4$ , so the perpendicular bisector is

y-5=-\frac34(x-1),\qquad 3x+4y=23.

Substitute $y=x+2$ : $3x+4(x+2)=23$ , giving $x=15/7$ and $y=29/7$ . Thus $P=(15/7,29/7)$ .

Check squared distances without introducing unnecessary square roots: $PA^2=(29/7)^2+(22/7)^2=1325/49$ , and $PB^2=(-13/7)^2+(-34/7)^2=1325/49$ .

A point described by two conditions must satisfy both. For a perpendicular bisector and another line, solve their equations simultaneously. For a point on an axis, use $x=0$ or $y=0$ before solving.

Divide a segment in a given ratio

If $AP:PB=m:n$ , where $m,n>0$ and $P$ lies between $A$ and $B$ , then $P$ is the fraction $m/(m+n)$ of the way from $A$ to $B$ . Apply this same fraction to both coordinate changes:

P=\left(x_1+\frac{m}{m+n}(x_2-x_1),\ y_1+\frac{m}{m+n}(y_2-y_1)\right).

Equivalently, $P=(\frac{nx_1+mx_2}{m+n},\frac{ny_1+my_2}{m+n})$ . The weight on $B$ is $m$ , because a larger $AP$ puts $P$ nearer $B$ . For a point on an extension, use a directed displacement from a known endpoint; do not apply the internal-division formula without checking the order of the points.

Example 3: a point two fifths along a segment

Given $A(1,2)$ , $B(11,7)$ and $AP:PB=2:3$ , find $P$ .

The total change is $(10,5)$ . Two fifths of it is $(4,2)$ , so $P=(1+4,2+2)=(5,4)$ . The remaining change to $B$ is $(6,3)$ , which is $3/2$ times $(4,2)$ . This confirms the ratio $2:3$ .

Explore: the same fraction in both coordinates

Move $P$ from $A(1,2)$ to $B(11,7)$ . Watch how its coordinates and the ratio change together. At the midpoint, both coordinate changes are halved.

Percentage of the way from A to B:40%

Use the slider or arrow keys. The endpoints are included so you can compare them with internal division.

Horizontal and vertical movements use the same fraction t. This keeps P on the straight line.

t = 0.4. P = (5, 4). AP : PB = 2 : 3.

Use properties of a shape to find a vertex

In a parallelogram the diagonals bisect each other. If $A,B,C,D$ are consecutive vertices, equating the midpoints of $AC$ and $BD$ gives $A+C=B+D$ , coordinate by coordinate. A rectangle also needs perpendicular adjacent sides. A sketch alone cannot establish either property.

Example 4: complete and verify a rectangle

The consecutive vertices of rectangle $ABCD$ are $A(1,1)$ , $B(5,3)$ , $C(4,5)$ . Find $D$ and its area.

Since the diagonals have the same midpoint, $D=A+C-B=(0,3)$ . The adjacent gradients are $m_{AB}=2/4=1/2$ and $m_{BC}=2/(-1)=-2$ , with product $-1$ , confirming the right angle.

Lengths are $AB=\sqrt{20}=2\sqrt5$ and $BC=\sqrt5$ . Therefore the area is $(2\sqrt5)(\sqrt5)=10$ square units. Both diagonal midpoints are $(5/2,3)$ .

Areas: use a perpendicular height or the shoelace rule

For a triangle with an easy horizontal or vertical base, use $\frac12\times\text{base}\times\text{perpendicular height}$ . For a general polygon, list its vertices in boundary order, clockwise or anticlockwise, and repeat the first vertex at the end. The shoelace rule is

\text{Area}=\frac12\left|\sum_{i=1}^{n}x_i y_{i+1}-\sum_{i=1}^{n}y_i x_{i+1}\right|,\qquad (x_{n+1},y_{n+1})=(x_1,y_1).

Each cross-product contributes a signed triangle area; contributions along internal edges cancel. The absolute value removes the sign caused by the direction of traversal. The rule gives the ordinary area for a simple polygon whose edges do not cross. A random ordering of vertices can create a different figure.

Example 5: the area of a quadrilateral

Find the area of $A(0,0), B(6,0), C(4,3), D(0,4)$ , in that boundary order.

\text{Area}=\frac12\left|(0\cdot0+6\cdot3+4\cdot4+0\cdot0)-(0\cdot6+0\cdot4+3\cdot0+4\cdot0)\right|=17.

The area is $17$ square units. Independently, diagonal $AC$ splits the figure into triangles of areas $\frac12(6)(3)=9$ and $\frac12(4)(4)=8$ ; these add to $17$ .

Boundary order matters. The two triangle areas give an independent check of the quadrilateral's area.

Example 6: an area condition can have two answers

Triangle $A(0,0), B(4,0), C(1,k)$ has area $6$ . Find $k$ .

The perpendicular height above or below the x-axis is $|k|$ , so $\frac12(4)|k|=6$ . Hence $|k|=3$ , giving $k=3$ or $k=-3$ . If the question additionally states that $C$ is above the x-axis, only $k=3$ remains. Both signs otherwise give area $6$ .

Common mistakes and useful checks

Giving a vertical line gradient zero. Zero gradient means horizontal. A vertical line has a constant x-coordinate.
Using a negative reciprocal without checking the case. If one line is horizontal, its perpendicular is vertical.
Confusing a perpendicular bisector with one point. It is a whole line of points equidistant from the two endpoints.
Reversing a ratio. Sketch which endpoint P should be nearer, then take the appropriate fraction of the displacement.
Assuming a quadrilateral's type. State the relevant equal lengths, gradients or diagonal midpoints.
Losing an area solution. Retain the absolute value until positional information determines the sign.

Practise coordinate geometry

Use the Secondary 3 coordinate geometry questions for these core methods. For mixed revision, use the Secondary 4 coordinate geometry questions. Continue with coordinate geometry of circles when the distance from a fixed centre is constant.