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Algebra · Topic 4 of 12

Straight line

Video lesson5 worked examples

Theory

Gradient

The gradient (m) defines the steepness of a line. The formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$ .

Equation of a Line

A straight line is usually written as $y = mx + c$ , where $m$ is the gradient and $c$ is the $y$ -intercept. When given the gradient and any point $(a, b)$ , use $y - b = m(x - a)$ instead.

The General Form

Lines are often given in the general form $ax + by + c = 0$ (for example $3x - 7y - 4 = 0$ ). You cannot read the gradient off this directly — you must rearrange it into $y = mx + c$ first, and then $m$ and the $y$ -intercept can be read off. This is one of the most frequently examined straight-line skills.

To find where any line crosses the axes: set $y = 0$ to find the $x$ -axis crossing, and set $x = 0$ to find the $y$ -axis crossing.

Special Lines

A horizontal line has gradient $0$ and equation $y = c$ . A vertical line has an undefined gradient and equation $x = k$ . Parallel lines have equal gradients.

The Golden Rule: to write the equation of a line you always need a gradient and a point. Given two points, find the gradient first, then use $y - b = m(x - a)$ . Given a line in general form, rearrange to $y = mx + c$ before reading off anything.

⚠️ Common Examiner Traps

Reading the gradient off the general form: in $3x - 7y - 4 = 0$ the gradient is not 3. You must rearrange into $y = mx + c$ first — this is the single most common mistake here.
Gradient formula order: subtract the coordinates in the same order on top and bottom. Flipping one gives the wrong sign.
Sign of a negative point: in $y - b = m(x - a)$ , a point with $a = -2$ gives $x - (-2) = x + 2$ .
Special gradients: a horizontal line has gradient $0$ (equation $y = c$ ); a vertical line has an undefined gradient (equation $x = k$ ).

Worked examples

Example 1

Finding the Gradient

Calculate the gradient of the line passing through (−2, 4) and (4, 16).

Step 1: Apply the formula: $m = \frac{16 - 4}{4 - (-2)}$ .

Answer: $m = \frac{12}{6} = 2$ .

Example 2

Finding the Equation

Find the equation of the line passing through (3, 5) with a gradient of −2.

Step 1: Substitute into $y - b = m(x - a)$ : $y - 5 = -2(x - 3)$ .

Step 2: Expand and rearrange: $y - 5 = -2x + 6 \rightarrow y = -2x + 11$ .

Example 3

🔗 Bringing it together

Find the equation of the line passing through the points $(1, 3)$ and $(4, 12)$ .

Step 1: Two points, no gradient given — so find the gradient first: $m = \dfrac{12 - 3}{4 - 1} = \dfrac{9}{3} = 3$ .

Step 2: Now use $y - b = m(x - a)$ with the gradient and either point. Taking $(1, 3)$ : $y - 3 = 3(x - 1)$ .

Step 3: Expand and rearrange:

y - 3 = 3x - 3 \implies y = 3x

Answer: $y = 3x$ .

Example 4

Gradient from the General Form

A straight line has equation $4x + 2y - 10 = 0$ . Find its gradient and the coordinates of its $y$ -intercept.

Step 1: You cannot read anything off yet — rearrange into $y = mx + c$ . Move the other terms to the right:

2y = -4x + 10

Step 2: Divide every term by 2 to leave $y$ on its own:

y = -2x + 5

Step 3: Now read off the values: $m = -2$ and $c = 5$ .

Answer: gradient $-2$ , $y$ -intercept $(0, 5)$ .

Example 5

🎯 Exam-style

A straight line has equation $3x + 5y - 15 = 0$ . Find the coordinates of the points where it crosses the $x$ -axis and the $y$ -axis.

Step 1: A line crosses the $x$ -axis where $y = 0$ . Substitute:

3x + 5(0) - 15 = 0 \implies 3x = 15 \implies x = 5

So it crosses the $x$ -axis at $(5, 0)$ .

Step 2: It crosses the $y$ -axis where $x = 0$ . Substitute:

3(0) + 5y - 15 = 0 \implies 5y = 15 \implies y = 3

So it crosses the $y$ -axis at $(0, 3)$ .

Answer: $(5, 0)$ on the $x$ -axis and $(0, 3)$ on the $y$ -axis.

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15 Straight Line questions with answers and video solutions