Coordinate geometry: results and formulae

This lesson is a quick revision of the main results and formulae of coordinate geometry, the kind asked for in one mark questions. It moves from the axes and quadrants through the distance formula to the section, midpoint, and centroid formulae.

Axes, quadrants and coordinates

The point where the $x$-axis and $y$-axis meet is the origin. The two axes divide the plane into four parts called quadrants.

A point has two coordinates: the $x$-coordinate (the abscissa) and the $y$-coordinate (the ordinate). Two ordered pairs are equal exactly when their parts match:

$(a, b) = (c, d) \iff a = c \text{ and } b = d.$

On the $x$-axis the $y$-coordinate of a point is $0$, so it has the form $(x, 0)$. On the $y$-axis the $x$-coordinate is $0$, so it has the form $(0, y)$. The origin itself is $(0, 0)$.

Signs of coordinates in each quadrant

The sign of each coordinate tells you which quadrant a point lies in.

• First quadrant: $(+, +)$, for example $(2, 5)$.
• Second quadrant: $(-, +)$, for example $(-3, 6)$.
• Third quadrant: $(-, -)$, for example $(-4, -6)$.
• Fourth quadrant: $(+, -)$, for example $(8, -2)$.

A point such as $(0, 3)$ lies on the $y$-axis and $(4, 0)$ lies on the $x$-axis.

The distance formula

The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is

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

The order of subtraction does not matter, since squaring removes the sign:

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

For the distance from the origin $O(0, 0)$ to a point $P(x, y)$ this reduces to

$OP = \sqrt{x^2 + y^2}.$

Identifying figures from side lengths

Using the distance formula to compare sides and diagonals, you can recognise different figures.

• Scalene triangle: all three sides are different.
• Equilateral triangle: all three sides are equal.
• Isosceles triangle: two sides are equal.
• Right triangle: the square of the longest side equals the sum of the squares of the other two sides.
• Isosceles right triangle: show it is both isosceles and right.
• Rectangle: opposite sides are equal and the diagonals are equal.
• Square: all sides are equal and the diagonals are equal.
• Rhombus: all sides are equal but the diagonals are not equal.
• Parallelogram: opposite sides are equal but the diagonals are not equal.
• Collinear points $A$, $B$, $C$: find $AB$, $BC$, and $CA$. If the longest of these equals the sum of the other two, the points are collinear.

The section formula and midpoint

If $P(x, y)$ divides the segment $AB$ internally in the ratio $m_1 : m_2$, where $A(x_1, y_1)$ and $B(x_2, y_2)$, then

$P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2},\ \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right).$

The midpoint of $AB$ is the special case of equal parts:

$M = \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right).$

The centroid of a triangle

The point where the medians of a triangle meet is the centroid, denoted $G$. The centroid divides each median in the ratio $2 : 1$ from the vertex, so $AG : GD = BG : GE = CG : GF = 2 : 1$.

For a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, the centroid is

$G = \left( \frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3} \right).$

Key takeaways

• A point on the $x$-axis is $(x, 0)$ and a point on the $y$-axis is $(0, y)$; the signs of the coordinates fix the quadrant.
• The distance between $A(x_1, y_1)$ and $B(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, and from this you can identify triangles and quadrilaterals.
• The section formula gives the point dividing $AB$ in ratio $m_1 : m_2$, and the centroid of a triangle is the average of its three vertices.