Points, Straight Lines and Shapes in 2D

Lessons

Senior 5 · Mathematics

Points, Straight Lines and Shapes in 2D

4 lessons

Lesson 1 of 4

Points and the equations of a line

25 min

From a sketch to equations

A surveyor who marks the corners of a plot does not draw them; she records them as coordinates, in metres east and north of a fixed point. Every question about the plot then becomes algebra: how long is a boundary, where is its midpoint, does a planned road cross it? In S3 you met the gradient and the equation $y = m x + c$ , and in S4 vectors in $\mathbb{R}^2$ and the scalar product. This unit joins the two. A line will be described by a point and a vector, which handles vertical lines as easily as any other and leads straight to angles, distances and, in S5 Unit 8, lines in three dimensions.

Important

Points in the plane (for $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ )

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, \qquad M_{AB} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).

Division point: the point $P$ on segment $A B$ with $A P : P B = m : n$ is $P = A + \frac{m}{m + n}\overrightarrow{AB}$ , that is

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

Centroid of triangle $A B C$ : $G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$ , where the three medians meet.

Worked example 1: length, midpoint and a division point

For $A (- 2, 5)$ and $B (4, - 3)$ , find $A B$ , the midpoint of $A B$ , and the point $P$ of $A B$ with $A P : P B = 1 : 3$ .

Solution $\overrightarrow{AB} = (6, -8)$ , so $AB = \sqrt{36 + 64} = 10$ . The midpoint is $\left(\frac{-2 + 4}{2}, \frac{5 - 3}{2}\right) = (1, 1)$ . $P$ is a quarter of the way from $A$ : $P = (-2, 5) + \frac{1}{4}(6, -8) = (-0.5, 3)$ . Check: $AP = \frac{1}{4} \times 10 = 2.5$ and $\sqrt{1.5^2 + 2^2} = 2.5$ . ✓

Definition

Direction vector of a line: any non-zero vector $\vec{u}$ parallel to the line.

Vector equation of the line through the point $A$ with direction $\vec{u}$ : a point $P$ lies on the line if and only if $\overrightarrow{AP} = t\vec{u}$ for some real number $t$ , that is

\overrightarrow{OP} = \overrightarrow{OA} + t\,\vec{u}, \qquad t \in \mathbb{R}.

Parameter: the variable $t$ . Parametric equations: the same equation written coordinate by coordinate. For $A (x_{A}, y_{A})$ and $\vec{u} = (u_1, u_2)$ :

x = x_A + t\,u_1, \qquad y = y_A + t\,u_2.

Example: the line through $(1, 2)$ with direction $(3, 1)$ is $x = 1 + 3 t$ , $y = 2 + t$ . Here $t = 2$ gives the point $(7, 4)$ .

The Cartesian equation and the normal vector

Eliminate $t$ from the parametric equations: $u_{2} (x - x_{A}) - u_{1} (y - y_{A}) = 0$ . This has the form

ax + by + c = 0, \qquad (a, b) \ne (0, 0),

the Cartesian (general) equation of the line. Read backwards, the equation $a x + b y + c = 0$ tells you two vectors at once:

$\vec{u} = (-b, a)$ is a direction vector;
$\vec{n} = (a, b)$ is a normal vector: it is perpendicular to the line, because $\vec{n}\cdot\vec{u} = -ab + ba = 0$ .

If $b \ne 0$ , the gradient is $m = -\frac{a}{b}$ ; if $b = 0$ , the line $a x + c = 0$ is vertical and has no gradient, but it still has the direction vector $(0, a)$ .

Figure 1. The line through

A (1, 2)

with direction

\vec{u} = (3, 1)

. Each value of the parameter

t

gives one point

(1 + 3 t, 2 + t)

. The normal vector

\vec{n} = (1, -3)

is perpendicular to

\vec{u}

, since

3 \times 1 + 1 \times (-3) = 0

Definition

Normal vector of a line: any non-zero vector perpendicular to the line. The line through $A (x_{A}, y_{A})$ with normal $\vec{n} = (a, b)$ is the set of points $P$ with $\vec{n}\cdot\overrightarrow{AP} = 0$ :

a (x - x_{A}) + b (y - y_{A}) = 0.

Example: the line through $(3, - 1)$ with normal $(4, - 3)$ is $4 (x - 3) - 3 (y + 1) = 0$ , that is $4 x - 3 y - 15 = 0$ .

Worked example 2: three forms of one line

Find the vector, parametric and Cartesian equations of the line through $A (1, 2)$ and $B (7, 4)$ . Do $C (10, 5)$ and $D (4, 4)$ lie on it?

Solution $\overrightarrow{AB} = (6, 2) = 2(3, 1)$ , so take $\vec{u} = (3, 1)$ .

Vector: $\overrightarrow{OP} = (1, 2) + t(3, 1)$ .
Parametric: $x = 1 + 3 t$ , $y = 2 + t$ .
Cartesian: from the second, $t = y - 2$ ; then $x = 1 + 3 (y - 2)$ , so $x - 3 y + 5 = 0$ . Check with $B$ : $7 - 12 + 5 = 0$ . ✓

$C$ : $10 - 15 + 5 = 0$ , so $C$ is on the line (it is the point $t = 3$ ). $D$ : $4 - 12 + 5 = -3 \ne 0$ , so $D$ is not.

Worked example 3: from Cartesian to parametric

Write parametric equations for $2 x + 5 y - 10 = 0$ , and give its gradient and a normal vector.

Solution A point: $y = 0$ gives $x = 5$ , so $(5, 0)$ is on the line. A direction vector is $(- b, a) = (- 5, 2)$ . So $x = 5 - 5 t$ , $y = 2 t$ . The gradient is $-\frac{2}{5}$ and a normal vector is $(2, 5)$ .

Warning

A line has many equations of each kind. Any point of the line and any non-zero multiple of the direction vector give a correct vector equation: $(1, 2) + t (3, 1)$ and $(7, 4) + s (- 6, - 2)$ describe the same line. So two answers that look different may both be right; test one by checking that its point satisfies the other's Cartesian equation and that the direction vectors are parallel.

Tip

From $a x + b y + c = 0$ at a glance: the normal is $(a, b)$ , the coefficients themselves; for a direction, swap them and change one sign: $(- b, a)$ or $(b, - a)$ .

ProfRobBob: Vector Form, Parametric Form, Cartesian Form of a Line 2D. Watch how one line through two given points is written in all three forms, and how the parameter is eliminated to reach the Cartesian equation.

✦

Practice

The last two are harder.

For $P (- 3, 4)$ and $Q (5, - 2)$ , find $P Q$ and the midpoint of $P Q$ .
Find the point dividing the segment from $A (2, - 1)$ to $B (10, 7)$ in the ratio $3 : 1$ .
Write the vector, parametric and Cartesian equations of the line through $(- 1, 3)$ with direction vector $(2, - 5)$ .
Find the Cartesian equation of the line $x = 4 - 2 t$ , $y = 1 + 3 t$ .
For $3 x - 4 y + 12 = 0$ , give a direction vector, a normal vector and the gradient.
Does $(7, - 2)$ lie on the line $x = 1 + 2 t$ , $y = 4 - 2 t$ ?
Find the equation of the line through $A (2, 3)$ with normal vector $(5, 1)$ .
For $A (1, 1)$ , $B (7, 3)$ , $C (4, 8)$ , find the centroid $G$ , and show that $G$ divides the median from $A$ in the ratio $2 : 1$ .

Show answersHide answers

$PQ = \sqrt{64 + 36} = 10$ ; midpoint $(1, 1)$ .
$A + \frac{3}{4}\overrightarrow{AB} = (2, -1) + \frac{3}{4}(8, 8) = (8, 5)$ .
$\overrightarrow{OP} = (-1, 3) + t(2, -5)$ ; $x = - 1 + 2 t$ , $y = 3 - 5 t$ ; $5 x + 2 y - 1 = 0$ .
$t = \frac{4 - x}{2}$ , so $y = 1 + \frac{3(4 - x)}{2}$ : $3 x + 2 y - 14 = 0$ .
Direction $(4, 3)$ , normal $(3, - 4)$ , gradient $\frac{3}{4}$ .
From $x$ : $t = 3$ ; then $y = 4 - 6 = - 2$ . Yes, it is the point $t = 3$ .
$5 (x - 2) + (y - 3) = 0$ : $5 x + y - 13 = 0$ .
$G = (4, 4)$ . The midpoint of $B C$ is $M (5.5, 5.5)$ . $\overrightarrow{AG} = (3, 3)$ and $\overrightarrow{GM} = (1.5, 1.5)$ , so $\overrightarrow{AG} = 2\overrightarrow{GM}$ : $A G : G M = 2 : 1$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S5 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Vector Form Parametric Form Cartesian Form of a Line 2 D, ProfRobBob, embedded from YouTube.

1From a sketch to equations

2The Cartesian equation and the normal vector

From a sketch to equations

The Cartesian equation and the normal vector