Distance, Midpoint and Gradient | IEB Grade 12 Maths

Q: Are the distance and midpoint formulae on the IEB formula sheet?

Yes, along with the gradient and line equation formulae. You still need to use them accurately and quickly.

Q: How do I find the angle of inclination of a line with a negative gradient?

Use $\tan^{-1}$ of the gradient, then add 180° to the negative answer so the angle lies between 0° and 180°.

Q: How do I prove three points are collinear?

Show that the gradient between the first two equals the gradient between two others, with a shared point.

Analytical geometry puts shapes on the Cartesian plane and proves things about them with algebra. This topic covers the tools every Paper 2 analytical geometry question uses: distance, midpoint, gradient, inclination and the equation of a line.

Distance and midpoint

For $A (x_{1}; y_{1})$ and $B (x_{2}; y_{2})$ :

$A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

$M_{A B} = (\frac{x _{1} + x _{2}}{2}; \frac{y _{1} + y _{2}}{2})$

Example. $A (- 2; 3)$ and $B (4; - 5)$ .

$A B = 6^{2} + (- 8)^{2} = 10 M = (1; - 1)$

If you know a midpoint and one endpoint, work backwards: if $M (1; - 1)$ is the midpoint of $A B$ and $A (- 2; 3)$ , then $\frac{- 2 + x}{2} = 1$ gives $x = 4$ .

Gradient, parallel and perpendicular lines

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

Parallel lines have equal gradients: $m_{1} = m_{2}$ .
Perpendicular lines have gradients with a product of $- 1$ : $m_{1} \times m_{2} = - 1$ .
A horizontal line has $m = 0$ . A vertical line has an undefined gradient.

Example. The line $2 x - y + 4 = 0$ has $m = 2$ . A perpendicular line has $m = - \frac{1}{2}$ .

Always rewrite a line in the form $y = m x + c$ before reading off the gradient.

Angle of inclination

The angle of inclination $θ$ is the angle a line makes with the positive $x$ -axis, measured anticlockwise, with $0° \leq θ < 180°$ .

$m = tan θ$

If $m$ is negative, the calculator gives a negative angle. Add 180°: $m = - 1$ gives $θ = - 45° + 180° = 135°$ .

The angle between two lines. Find each inclination, then subtract. Use exterior angle of a triangle reasoning on the diagram.

Example. Line $A B$ has $m = 1, 5$ , so $θ_{1} \approx 56, 31°$ . Line $C D$ has $θ_{2} = 135°$ . The angle between them is $135° - 56, 31° = 78, 69°$ .

Equation of a straight line

Use either form:

Gradient-point form: $y - y_{1} = m (x - x_{1})$
Gradient-intercept form: $y = m x + c$ , substituting a point to find $c$

Example. Line through $P (3; - 1)$ perpendicular to $y = 2 x + 4$ :

$m = - \frac{1}{2} \Rightarrow y + 1 = - \frac{1}{2} (x - 3) \Rightarrow y = - \frac{1}{2} x + \frac{1}{2}$

Equations of special lines: the median passes through a vertex and the midpoint of the opposite side; the altitude passes through a vertex perpendicular to the opposite side; the perpendicular bisector passes through the midpoint of a side, perpendicular to it.

Collinearity

Points are collinear if they lie on one straight line. Show this by proving two gradients are equal and share a point:

$m_{A B} = m_{B C} \Rightarrow A, B, C are collinear$

Example. $A (1; 2)$ , $B (5; k)$ and $C (7; 8)$ are collinear. Then $m_{A B} = m_{A C}$ :

$\frac{k - 2}{4} = \frac{6}{6} = 1 \Rightarrow k = 6$

Key terms

Gradient: The steepness of a line, $\frac{Δ y}{Δ x}$ .
Angle of inclination: The angle between a line and the positive $x$ -axis, measured anticlockwise.
Collinear: Lying on the same straight line.
Perpendicular bisector: A line through the midpoint of a segment at right angles to it.
Median: A line from a vertex of a triangle to the midpoint of the opposite side.

Common mistakes

Subtracting coordinates in a different order in the numerator and denominator of the gradient.
Giving a negative angle of inclination instead of adding 180°.
Forgetting to rewrite a line as $y = m x + c$ before reading the gradient.
Using $m_{1} \times m_{2} = 1$ instead of $- 1$ for perpendicular lines.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

knowledge

Given $A (- 2; 3)$ and $B (4; - 5)$ . Calculate:

(a) the length of $A B$ (2)

(b) the coordinates of the midpoint of $A B$ (2)

Worked solution

Hidden - try the question first

Solution

(a) $A B = (4 + 2)^{2} + (- 5 - 3)^{2} = 36 + 64$
$= 10$
(b) $M = (\frac{- 2 + 4}{2}; \frac{3 - 5}{2})$
$= (1; - 1)$

Question 2

5 marks total

routine

Determine the equation of the line through $P (3; - 1)$ that is perpendicular to $2 x - y + 4 = 0$ .

Worked solution

Hidden - try the question first

Solution

$y = 2 x + 4$ , so $m = 2$
$m_{⊥} = - \frac{1}{2}$
$y + 1 = - \frac{1}{2} (x - 3)$
$y = - \frac{1}{2} x + \frac{3}{2} - 1$
$y = - \frac{1}{2} x + \frac{1}{2}$

Question 3

6 marks total

complex

Line $A B$ passes through $A (- 1; - 2)$ and $B (3; 4)$ . Line $C D$ has an angle of inclination of 135° and cuts $A B$ .

(a) Calculate the angle of inclination of $A B$ . (2)

(b) Determine the gradient of $C D$ . (1)

Worked solution

Hidden - try the question first

Solution

(a) $A B = (4 + 2)^{2} + (- 5 - 3)^{2} = 36 + 64$
$θ = tan^{- 1} (1, 5) \approx 56, 31°$
(b) $m_{C D} = tan 135° = - 1$
(c) In the triangle formed with the $x$ -axis, the exterior angle $135°$ equals the sum of the interior opposite angles
Angle between the lines $= 135° - 56, 31°$
$= 78, 69°$

Question 4

8 marks total

problem-solving

$A (1; 2)$ , $B (5; k)$ and $C (7; 8)$ are collinear.

(a) Determine the value of $k$ . (3)

(b) $D$ lies on the $y$ -axis with $A D ⊥ A C$ . Determine the coordinates of $D$ . (3)

Worked solution

Hidden - try the question first

Solution

(a) $A B = (4 + 2)^{2} + (- 5 - 3)^{2} = 36 + 64$
$m_{A B} = \frac{k - 2}{4} = 1$
$k = 6$
(b) $m_{A D} = - 1$
$y - 2 = - (x - 1) \Rightarrow y = - x + 3$
At $x = 0$ : $D (0; 3)$
(c) $D C = (7 - 0)^{2} + (8 - 3)^{2}$
$= 74$ units

FAQs

Are the distance and midpoint formulae on the IEB formula sheet?

Yes, along with the gradient and line equation formulae. You still need to use them accurately and quickly.

How do I find the angle of inclination of a line with a negative gradient?

Use $tan^{- 1}$ of the gradient, then add 180° to the negative answer so the angle lies between 0° and 180°.

How do I prove three points are collinear?

Show that the gradient between the first two equals the gradient between two others, with a shared point.

Distance, midpoint, gradient and inclination

Distance and midpoint

Gradient, parallel and perpendicular lines

Angle of inclination

Equation of a straight line

Collinearity

Key terms

Common mistakes

Practice questions

FAQs

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