Equation of a Straight Line

Q: How do I find where two lines meet?

Set their equations equal and solve for $x$, then substitute to find $y$.

Q: Do I need $y = mx + c$ form?

Give the equation in the form the question asks for. $y = mx + c$ is the usual final form.

Finding the equation of a line is the most common skill in analytical geometry. In Grade 11 you use it for sides, medians, altitudes and perpendicular bisectors of triangles, and then to find where lines meet.

Gradient-point and two-point forms

Gradient-point form: if you know $m$ and a point $(x_{1}; y_{1})$ :

$y - y_{1} = m (x - x_{1})$

Two-point form: find $m$ from the two points first, then use either point.

Example. Through $A (- 1; 4)$ and $B (3; - 4)$ :

$m = \frac{- 4 - 4}{3 + 1} = - 2 \Rightarrow y - 4 = - 2 (x + 1) \Rightarrow y = - 2 x + 2$

Check with the other point: $- 2 (3) + 2 = - 4$ ✓

Parallel and perpendicular lines

Parallel: $m_{1} = m_{2}$
Perpendicular: $m_{1} \times m_{2} = - 1$

Perpendicular bisector of $A B$ . It passes through the midpoint of $A B$ and is perpendicular to $A B$ .

For $A (- 1; 4)$ and $B (3; - 4)$ : the midpoint is $(1; 0)$ and $m_{⊥} = \frac{1}{2}$ , so

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

Altitude from $C$ . Perpendicular to the opposite side and through $C$ . Median from $C$ . Through $C$ and the midpoint of the opposite side.

Distance and midpoint revision

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} M = (\frac{x _{1} + x _{2}}{2}; \frac{y _{1} + y _{2}}{2})$

Example. $A B$ for $A (- 1; 4)$ and $B (3; - 4)$ :

$A B = 16 + 64 = 80 = 45$

Area of a triangle. Find a base length and the perpendicular height (the length of the altitude), then use $\frac{1}{2} \times$ base $\times$ height.

Key terms

Gradient-point form: $y - y_{1} = m (x - x_{1})$ .
Perpendicular bisector: A line through the midpoint of a segment, perpendicular to it.
Altitude: A line from a vertex perpendicular to the opposite side.
Median: A line from a vertex to the midpoint of the opposite side.

Common mistakes

Mixing up the median, altitude and perpendicular bisector.
Using the original gradient instead of the perpendicular gradient.
Sign errors when substituting negative coordinates.
Not simplifying surd lengths.

Practice questions

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

Question 1

3 marks total

knowledge

Determine the equation of the line with gradient 3 that passes through $(2; - 1)$ .

Worked solution

Hidden - try the question first

Solution

$y + 1 = 3 (x - 2)$
$y = 3 x - 6 - 1$
$y = 3 x - 7$

Question 2

5 marks total

routine

Determine the equation of the line through $A (- 1; 4)$ and $B (3; - 4)$ .

Worked solution

Hidden - try the question first

Solution

$y + 1 = 3 (x - 2)$
$= - 2$
$y - 4 = - 2 (x + 1)$
$y = - 2 x - 2 + 4$
$y = - 2 x + 2$

Question 3

6 marks total

complex

Determine the equation of the perpendicular bisector of $A B$ , with $A (- 1; 4)$ and $B (3; - 4)$ .

Worked solution

Hidden - try the question first

Solution

Midpoint $M = (\frac{- 1 + 3}{2}; \frac{4 - 4}{2}) = (1; 0)$
$m_{A B} = - 2$
$m_{⊥} = \frac{1}{2}$
$y - 0 = \frac{1}{2} (x - 1)$
$y = \frac{1}{2} x - \frac{1}{2}$

Question 4

9 marks total

problem-solving

$△ A B C$ has vertices $A (- 1; 4)$ , $B (3; - 4)$ and $C (5; 6)$ .

(a) Calculate the length of $A B$ in simplest surd form. (2)

(b) Determine the equation of the altitude from $C$ to $A B$ . (3)

(d) Calculate the area of $△ A B C$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $M = (\frac{- 1 + 3}{2}; \frac{4 - 4}{2}) = (1; 0)$
$= 45$
(b) $m_{A B} = - 2$ , so the altitude has $m = \frac{1}{2}$
$y - 6 = \frac{1}{2} (x - 5)$
$y = \frac{1}{2} x + \frac{7}{2}$
(c) $- 2 x + 2 = \frac{1}{2} x + \frac{7}{2} \Rightarrow x = - \frac{3}{5}$
$D (- \frac{3}{5}; \frac{16}{5})$
(d) $C D = (\frac{28}{5})^{2} + (\frac{14}{5})^{2} = \frac{14}{5}$
Area $= \frac{1}{2} \times 45 \times \frac{14}{5} = 28$ square units

FAQs

What is the difference between a median and an altitude?

A median goes to the midpoint of the opposite side. An altitude meets the opposite side at a right angle.

How do I find where two lines meet?

Set their equations equal and solve for $x$ , then substitute to find $y$ .

Do I need $y = m x + c$ form?

Give the equation in the form the question asks for. $y = m x + c$ is the usual final form.

Gradient-point and two-point forms

Parallel and perpendicular lines

Distance and midpoint revision

Key terms

Common mistakes

Practice questions

FAQs

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