C3.5; E3.5 - Equations of straight lines

C3.5; E3.5 - Equations of straight lines

A line's equation records its gradient and where it crosses an axis. You will interpret these features and find a fully simplified equation from a graph, including horizontal and vertical lines, for C3.5 and E3.5.

Gradient and intercept in an equation

Can two numbers describe an entire sloping line? In y=3x+2y=3x+2, one number tells you how fast the line rises; the other tells you its starting height on the vertical axis. Together they locate every point.

Try two inputs. At x=0x=0, the equation gives y=2y=2. At x=1x=1, it gives y=5y=5. Increasing xx by one raises yy by three. The same happens wherever you start, because the added two stays fixed.

Gradient and intercept form

For a non-vertical straight line,

y=mx+c,y=mx+c,

where mm is the gradient and cc is the yy-intercept, the value of yy where the line meets the yy-axis. That crossing point is (0,c)(0,c). Not given in the List of formulas: learn it.

The intercept is a value, so “the yy-intercept is 22” and “the line crosses at (0,2)(0,2)” describe the same feature. It is not generally where the line crosses the horizontal axis. Negative gradients and negative intercepts are allowed.

Worked example 1: reading both features

Question. State the gradient and yy-intercept of y=3x+2y=3x+2.

Plan. Match the coefficient of xx and the constant to y=mx+cy=mx+c.

y=3x+2already in the required formm=3coefficient of xc=2constant term\begin{aligned} y&=3x+2 && \text{already in the required form}\\ m&=3 && \text{coefficient of }x\\ c&=2 && \text{constant term} \end{aligned}

Answer. Gradient 33; yy-intercept 22, at (0,2)(0,2).

Check. Inputs 00 and 11 give outputs 22 and 55: the crossing is at two and a run of one produces a rise of three.

Change only the constant's sign. The line now crosses below the origin, but its gradient stays the same.

Worked example 2: a negative intercept

Question. State the gradient and yy-intercept of y=3x−2y=3x-2.

Plan. Keep the minus sign attached to the constant.

y=3x+(−2)show the signed constantm=3coefficient of xc=−2constant term\begin{aligned} y&=3x+(-2) && \text{show the signed constant}\\ m&=3 && \text{coefficient of }x\\ c&=-2 && \text{constant term} \end{aligned}

Answer. Gradient 33; yy-intercept −2-2, at (0,−2)(0,-2).

Check. At x=0x=0, y=−2y=-2; at x=1x=1, y=1y=1. The rise is still three for a run of one.

Simplify the right side before reading the features: y=2x+3+x−5y=2x+3+x-5 becomes y=3x−2y=3x-2. Writing y=−x+4y=-x+4 means gradient −1-1; a missing written coefficient does not mean zero.

Watch out

In y=3x−2y=3x-2, the gradient is 33, not −2-2. Substitute x=0x=0 to identify the intercept, rather than choosing whichever number looks nearest the equals sign.

Finding an equation from a graph

To turn a graph into an equation, read its yy-intercept and measure its gradient from a grid triangle. Put those values into y=mx+cy=mx+c. Then check another point on the graph. Reading the intercept alone cannot distinguish lines that cross the same point in different directions.

[DIAGRAM: asset_name: Two lines with the same intercept; asset_slug: graph_equations; caption: Both lines cross at negative three; changing the direction changes the gradient.; description: Two numbered Cartesian grids use one x-unit per horizontal division and two y-units per vertical division. The left line rises through zero comma negative three and two comma one; its run is two and rise four. The right line falls through zero comma negative three and two comma negative seven; its run is two and rise negative four. A third point is marked on each line for checking.]
Both lines cross at negative three

Worked example 3: a rising graph

Question. Find the equation of the line in the left grid.

Plan. Read the crossing value and use the displayed triangle for the gradient.

c=−3read the y-axis crossingm=42=2rise four; run twoy=2x−3substitute into y=mx+c\begin{aligned} c&=-3 && \text{read the y-axis crossing}\\ m&=\frac42=2 && \text{rise four; run two}\\ y&=2x-3 && \text{substitute into }y=mx+c \end{aligned}

Answer. y=2x−3y=2x-3.

Check. The separate graph point (−1,−5)(-1,-5) gives 2(−1)−3=−52(-1)-3=-5, so the equation agrees with it.

Change only the gradient's sign, keeping the intercept at negative three. The second graph falls to the right.

Worked example 4: a falling graph

Question. Find the equation of the line in the right grid.

Plan. Read the same intercept, but give the downward rise a negative sign.

c=−3read the crossingm=−42=−2down four; right twoy=−2x−3combine gradient and intercept\begin{aligned} c&=-3 && \text{read the crossing}\\ m&=\frac{-4}{2}=-2 && \text{down four; right two}\\ y&=-2x-3 && \text{combine gradient and intercept} \end{aligned}

Answer. y=−2x−3y=-2x-3.

Check. At the separate marked point (−1,−1)(-1,-1), the equation gives −2(−1)−3=−1-2(-1)-3=-1, matching the graph.

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