C4.4; E4.4 - Equations of straight lines | Cambridge IGCSE International Mathematics

A line crosses the $y$ -axis at 1 and rises 2 units for every unit to the right. How can one equation describe every point on it? C4.4 and E4.4 connect a straight-line graph's gradient and intercept to its equation, including horizontal and vertical lines.

Slope-intercept form

Start on the $y$ -axis at height 1. After moving $x$ units horizontally along a line with gradient 2, the height changes by $2x$ . Its height is therefore $y=2x+1$ . This describes every point, rather than only the starting point.

Gradient and y-intercept

For the non-vertical straight line $y=mx+c$ , $m$ is its gradient and $c$ is its $y$ -intercept, the value of $y$ where $x=0$ . The intercept point is $(0,c)$ . This equation form is not on the supplied formula list for Papers 1–4: learn it.

The expression “slope-intercept form” names the same form: slope is another word for gradient. Multiplying $x$ by $m$ accounts for the change from the starting height $c$ . When $x=0$ , the $mx$ term vanishes, leaving $y=c$ . When $x$ increases by 1, $y$ changes by $m$ .

Worked example 1: identifying a positive gradient

Question. Write down the gradient and $y$ -intercept of $y=3x-2$ .

Plan. Compare the coefficient of $x$ and the constant term with $y=mx+c$ .

\begin{aligned} y&=3x+(-2)\\ m&=3\\ c&=-2 \end{aligned}

Answer. Gradient 3 and $y$ -intercept $-2$ , at the point $(0,-2)$ .

Check. $x=0$ gives $y=-2$ ; $x=1$ gives $y=1$ . A unit increase in $x$ produces a rise of $1-(-2)=3$ .

Keep the intercept but change the coefficient of $x$ . The new negative fractional gradient changes both direction and steepness.

Worked example 2: identifying a negative fractional gradient

Question. Write down the gradient and $y$ -intercept of $y=-\frac12x-2$ .

Plan. Include the minus sign in the coefficient and in the constant term.

\begin{aligned} y&=\left(-\frac12\right)x+(-2)\\ m&=-\frac12\\ c&=-2 \end{aligned}

Answer. Gradient $-\frac12$ and $y$ -intercept $-2$ , at $(0,-2)$ .

Check. At $x=0$ , $y=-2$ ; at $x=2$ , $y=-3$ . The change is $-1$ over a run of 2, giving $-1/2$ .

An omitted coefficient of $x$ means 1: $y=x+4$ has gradient 1. In $y=-x+4$ , the coefficient is $-1$ . A line through the origin has $c=0$ , so $y=5x$ is already simplified.

Watch out

The $y$ -intercept of $y=3x-2$ is $-2$ , not 2 and not 3. The point $(0,-2)$ is on the $y$ -axis; $(-2,0)$ would be on the $x$ -axis.

Compare a fresh equation with the same form.

Equation from a graph

To obtain an equation from a grid, read where the line crosses the $y$ -axis and calculate its gradient using a clear triangle. Put these two values into $y=mx+c$ , then simplify. At Core, the grid supplies the gradient information; you do not need a coordinate-only gradient formula.

[DIAGRAM: asset_name: Equations from three straight-line grids; asset_slug: line_equation_grids; caption: Read the intercept at x equals zero and the gradient from a matching rise and run.; description: Three accurate Cartesian unit grids. Example 1 shows a line through zero comma one and two comma five, with those clear points marked. Example 2 shows a falling line through zero comma one and four comma minus one. Your turn shows a separate line through zero comma three and two comma minus one. The lines are not labelled with their equations or gradients.]
Read the intercept at x equals zero and the gradient from a matching rise and run.

Worked example 3: a positive gradient from the grid

Question. Find the equation of the line in the first panel.

Plan. Read the $y$ -intercept, then use the marked points to find the gradient.

\begin{aligned} c&=1 &&\text{the crossing at }x=0\\ m&=\frac42=2 &&\text{rise 4, run 2}\\ y&=2x+1 &&\text{substitute into }y=mx+c \end{aligned}

Answer. $y=2x+1$ , answering the opening question.

Check. The marked point $(2,5)$ satisfies $5=2(2)+1$ . The intercept $(0,1)$ also satisfies the equation.

The second grid keeps the same intercept but changes the gradient to a falling fraction. Simplify the ratio before writing the equation.

Worked example 4: a negative gradient from the grid

Question. Find the equation of the line in the second panel.

Plan. Use its intercept and the signed change from the left marked point to the right one.

\begin{aligned} c&=1\\ m&=\frac{-2}{4}=-\frac12 &&\text{fall 2, run 4}\\ y&=-\frac12x+1 \end{aligned}

Answer. $y=-\frac12x+1$ .

Check. At $x=4$ , the equation gives $y=-2+1=-1$ , agreeing with the right-hand marked point. At $x=0$ , it gives 1.

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C4.4; E4.4 - Equations of straight lines

Slope-intercept form

Equation from a graph

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