Coordinate Geometry — chapter notes

The distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$ is

It is Pythagoras' theorem in the right triangle with legs $|x_2 - x_1|$ (horizontal) and $|y_2 - y_1|$ (vertical). Special cases: from the origin, $\sqrt{x^2 + y^2}$; on a horizontal line, $|x_2 - x_1|$; on a vertical line, $|y_2 - y_1|$.

Example: $A(-3, 2)$, $B(5, -4)$: $AB = \sqrt{8^2 + 6^2} = 10$. Keep negative coordinates in brackets: $5 - (-3) = 8$.

Unknown coordinate: square both sides of the distance condition. Expect two answers: $PA = 10$ with $A(1, -1)$ and $P(x, 5)$ gives $(x - 1)^2 = 64$, so $x = 9$ or $x = -7$.

Equidistant point: set the squared distances equal; the squared terms cancel, leaving a linear equation.

• On the $x$-axis take $P(x, 0)$; on the $y$-axis take $P(0, y)$.
• Points equidistant from $A$ and $B$ satisfy a linear relation in $x$ and $y$ (e.g. $3x = 2y$ for $A(5, 1)$, $B(-1, 5)$): they form the perpendicular bisector of $AB$, which passes through the midpoint.

Collinear: $A$, $B$, $C$ lie on one line exactly when the largest of $AB$, $BC$, $CA$ equals the sum of the other two. (The middle point is the one not on the longest distance.) When decimals are close, compare exactly: $\sqrt{13} + \sqrt{18}$ is not $\sqrt{61}$.

Triangles: isosceles (two sides equal), equilateral (three), right-angled ($c^2 = a^2 + b^2$ for the longest side $c$, by the converse of Pythagoras' theorem).

Quadrilaterals (vertices in order): rhombus — four equal sides; square — four equal sides and equal diagonals; rectangle — opposite sides equal and equal diagonals; parallelogram — both pairs of opposite sides equal.

Not in the 2026–27 syllabus: area of a triangle from coordinates. Use distances for collinearity.

$P$ divides $AB$ internally in the ratio $m_1 : m_2$ when $P$ lies between $A$ and $B$ and $AP : PB = m_1 : m_2$. Then

Why: drawing horizontal and vertical lines through $A$, $P$, $B$ gives two right triangles with equal angles, which are similar (AA); their legs are in the ratio $m_1 : m_2$.

Pairing: $m_1$ (next to $A$) multiplies $B$'s coordinates. With $A(-2, 5)$, $B(8, -5)$, $3 : 2$: $P = (4, -1)$.

Fractions of the way: "one-third of the way from $A$" is $1 : 2$; "three-quarters of the way" is $3 : 1$.

Midpoint ($1 : 1$): $M\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$. Backwards: $B = (2x_M - x_1, 2y_M - y_1)$.

Trisection: the two points that cut $AB$ into three equal parts divide it in $1 : 2$ and $2 : 1$. For $A(2, -3)$, $B(11, 9)$: $(5, 1)$ and $(8, 5)$. Four equal parts: $1 : 3$, $1 : 1$, $3 : 1$.

Parallelogram $ABCD$: the diagonals $AC$ and $BD$ bisect each other, so they have the same midpoint; this finds a missing vertex ($D = A + C - B$ coordinate by coordinate).

Medians: the midpoint $D$ of $BC$ gives the median $AD$; the point dividing $AD$ in $2 : 1$ is the same for all three medians.

Write the unknown ratio as $k : 1$. Then $P = \left(\dfrac{kx_2 + x_1}{k + 1}, \dfrac{ky_2 + y_1}{k + 1}\right)$.

1. Equate the coordinate you know to its formula and solve for $k$.
2. Write $k : 1$ in whole numbers.
3. Use the other coordinate to find the point, or to check that $P$ is on $AB$.

Axes: the $x$-axis has $y = 0$; the $y$-axis has $x = 0$. The $x$-axis cuts the segment joining $A(1, -4)$ and $B(6, 6)$ in $2 : 3$ at $(3, 0)$. An axis cuts a segment internally only if the ends lie on opposite sides of it.