IB Applications and Interpretation SL · 2D geometry · Section 2 of 5

Gradient and equations of a line

3 worked examples · By Álvaro López

Key idea

A line is determined by two pieces of information: two points, or a point and its gradient. The gradient tells you how much the line rises or falls for each unit you move to the right, and with it you can write the equation in three equivalent forms. In the IB exam it appears when modelling linear situations (tariffs, paths, ramps), when finding where two lines cross, and when calculating shapes formed with the axes.

The gradient

When you travel along a line from left to right, your position changes in two directions: you move horizontally and you go up or down vertically. The gradient compares both changes.

[FIGURA mai-t4-a2-pendiente: Cartesian plane with x and y axes and a soft grid. A line that rises to the right passes through the points A(1, 2) and B(5, 5), marked with dots and labelled "A(x₁, y₁)" and "B(x₂, y₂)". From A a dashed horizontal segment runs to the vertical through B, labelled "x₂ − x₁ (run)"; from there a dashed vertical segment runs up to B, labelled "y₂ − y₁ (rise)". At the point where the line meets the x-axis, the angle θ between the line and the positive direction of the x-axis is marked. On the right, a small box with a line that falls (through C(1, 5) and D(5, 2)) with its run and fall marked, labelled "negative gradient"; the first drawing is labelled "positive gradient".]

The sign of $m$ tells you what the line is like:

$m > 0$ : the line rises as you move to the right.
$m < 0$ : the line falls as you move to the right.
$m = 0$ : the line is horizontal (there is no vertical change).
If $x_{1} = x_{2}$ , the denominator is $0$ : the line is vertical and its gradient is not defined.

The larger $∣ m ∣$ is, the steeper the line. A gradient $m = 2$ means that for every unit you move forward, you go up two.

Gradient and angle

If $θ$ is the acute angle that the line makes with the horizontal axis, the right-angled triangle in the figure has legs $y_{2} - y_{1}$ and $x_{2} - x_{1}$ . By the tangent ratio,

This relation links this topic with trigonometry: a gradient of $1$ corresponds to $4 5^{\circ}$ , and a large gradient to an angle close to $9 0^{\circ}$ .

The equations of a line

The same line can be written in several ways. Each form is useful for something different.

Point-gradient form

Suppose the line has gradient $m$ and passes through $A (x_{1}, y_{1})$ . If $P (x, y)$ is any other point on the line, the gradient between $A$ and $P$ is also $m$ : $\frac{y - y _{1}}{x - x _{1}} = m$ Clearing the denominator gives the following equation.

This is the form you use to construct a line when you know a point and the gradient, or two points (you calculate $m$ first and then use either of the two points).

Explicit form

If you rearrange the point-gradient form to make $y$ the subject, you get $y = m x + (y_{1} - m x_{1})$ The bracket is a constant: it is the value of $y$ when $x = 0$ .

It is the most convenient form for reading a line, sketching it or interpreting it as a model: $c$ is the initial value and $m$ is how much $y$ changes for each unit of $x$ .

General form

If you move all the terms to one side, the equation takes the form $a x + b y + d = 0$ with $a$ , $b$ and $d$ real numbers and $a$ , $b$ not both zero. This is the general (or standard) form. Exam questions often ask for “the form $a x + b y + d = 0$ , where $a$ , $b$ and $d$ are integers”: multiply the whole equation by the common denominator to remove the fractions.

To change from general form to explicit form, rearrange for $y$ : $b y = - a x - d \Rightarrow y = - \frac{a}{b} x - \frac{d}{b}$ From this you can read that the gradient is $m = - \frac{a}{b}$ and the y-intercept is $c = - \frac{d}{b}$ (if $b \neq = 0$ ).

Horizontal and vertical lines

[FIGURA mai-t4-a2-horizontal-vertical: two small Cartesian planes side by side, with axes graduated from -5 to 5 on x and from -4 to 4 on y. In the left one, a thick horizontal line at height y = -3, with the label "y = -3" above it and the point (2, -3) marked. In the right one, a thick vertical line at x = 4, with the label "x = 4" beside it and the point (4, 1) marked.]

Intercepts with the axes

[FIGURA mai-t4-a2-cortes-ejes: Cartesian plane with axes graduated from -2 to 7 on x and from -5 to 3 on y. A thick line that cuts the x-axis at the point (6, 0) and the y-axis at the point (0, -3). Both points marked with large dots and labelled "(6, 0)" and "(0, -3)". Beside the line, the label "x − 2y − 6 = 0". The triangle formed by the line and the positive part of the x-axis and the negative part of the y-axis is lightly shaded.]

Intercepts with the axes are the easiest points of a line to calculate and, in addition, they are enough to draw it: two points determine a single line.

Intersection of two lines

Two lines with different gradients meet at exactly one point. That point is the only one that satisfies both equations at once, so its coordinates are the solution of the system formed by both equations: ${a_{1} x + b_{1} y + d_{1} a_{2} x + b_{2} y + d_{2} = 0 = 0$ If one of the lines is in explicit form, the quickest method is to substitute $y$ into the other. If not, use elimination: multiply the equations so that the coefficients of one unknown match, and subtract.

If the gradients are equal and the y-intercepts are different, the lines are parallel and do not meet: the system has no solution.
If the gradients and the y-intercepts are equal, it is the same line: there are infinitely many solutions.

[FIGURA mai-t4-a2-interseccion: Cartesian plane with axes graduated from -2 to 8 on both x and y. Two lines: line r (blue) with equation "y = x/2 + 1", passing through (0, 1) and (4, 3); line s (orange) with equation "y = -x + 7", passing through (0, 7) and (7, 0). They meet at the point P(4, 3), marked with a large dot and labelled "P(4, 3)". Dashed lines from P to the axes, with the values 4 and 3 marked.]

Worked examples

Example 1 Basic

A path across a farm

On the plan of a farm, a straight track joins the farmhouse $A (2, 1)$ with the well $B (6, 4)$ (units of hundreds of metres).

a) Calculate the gradient of the track.

b) Find the acute angle that the track makes with the horizontal axis, to one decimal place.

c) Write the equation of the line containing the track in the form $a x + b y + d = 0$ , with $a$ , $b$ and $d$ integers and $a > 0$ .

Solution

a) $m = \frac{4 - 1}{6 - 2} = \frac{3}{4}$ .

b) Since $m = tan θ$ , $θ = tan^{- 1} \frac{3}{4} = 36. 9^{\circ}$ .

c) Point-gradient form using the point $A (2, 1)$ : $y - 1 = \frac{3}{4} (x - 2)$ We multiply by $4$ to remove the fraction: $4 y - 4 = 3 x - 6 \Rightarrow 3 x - 4 y - 2 = 0$ Check with $B (6, 4)$ : $3 \cdot 6 - 4 \cdot 4 - 2 = 18 - 16 - 2 = 0$ . It is satisfied.

Example 2 Medium

Bicycle hire

A shop hires out bicycles with a tariff that combines a fixed fee and a price per hour. Hiring a bicycle for $2$ hours costs $11$ euros and for $6$ hours costs $23$ euros. The cost $C$ (in euros) is a linear function of the time $t$ (in hours).

a) Find the gradient and the y-intercept of the line $C = m t + c$ that models the cost, and say what each one means.

b) How many hours can be hired for $32$ euros?

c) How much does it cost to hire the bicycle for $4.5$ hours?

Solution

a) The points on the line are $(2, 11)$ and $(6, 23)$ : $m = \frac{23 - 11}{6 - 2} = \frac{12}{4} = 3$ With the point $(2, 11)$ : $C - 11 = 3 (t - 2)$ , that is, $C = 3 t + 5$ . So $m = 3$ and $c = 5$ .

Meaning: $m = 3$ is the price per hour (3 euros for each extra hour); $c = 5$ is the fixed fee (what is paid even if the time were zero).

b) We solve $32 = 3 t + 5$ : $3 t = 27 \Rightarrow t = 9 hours$

c) $C = 3 \cdot 4.5 + 5 = 18.5$ euros.

Example 3 Hard

Two paths that cross

On the map of a park (units of 1 km), a path $L_{1}$ passes through the points $A (- 2, 1)$ and $B (4, 4)$ . Another path $L_{2}$ follows the line with equation $3 x + 2 y - 16 = 0$ .

a) Find the equation of $L_{1}$ in the form $a x + b y + d = 0$ with integer coefficients and $a > 0$ .

b) Find the points where $L_{2}$ cuts the $x$ -axis and the $y$ -axis.

c) Find the coordinates of the point where the two paths cross.

d) The line $L_{2}$ and the coordinate axes enclose a triangular region. Calculate its area in km $^{2}$ , to one decimal place.

Solution

a) Gradient: $m = \frac{4 - 1}{4 - ( - 2 )} = \frac{3}{6} = \frac{1}{2}$ . With the point $A (- 2, 1)$ : $y - 1 = \frac{1}{2} (x + 2) \Rightarrow 2 y - 2 = x + 2 \Rightarrow x - 2 y + 4 = 0$

b) With the x-axis ( $y = 0$ ): $3 x - 16 = 0$ , so $x = \frac{16}{3}$ and the point is $(\frac{16}{3}, 0)$ .

With the y-axis ( $x = 0$ ): $2 y - 16 = 0$ , so $y = 8$ and the point is $(0, 8)$ .

c) We solve the system: ${x - 2 y + 4 3 x + 2 y - 16 = 0 = 0$ Adding the two equations eliminates $y$ : $4 x - 12 = 0$ , so $x = 3$ . In the first, $3 - 2 y + 4 = 0$ , from which $y = \frac{7}{2}$ . The crossing point is $P (3, \frac{7}{2})$ . (Using the GDC, solving the system gives $x = 3$ , $y = 3.5$ .)

d) The triangle has vertices $O (0, 0)$ , $(\frac{16}{3}, 0)$ and $(0, 8)$ . It is right-angled at the origin, with legs $\frac{16}{3}$ and $8$ : $A = \frac{1}{2} \cdot \frac{16}{3} \cdot 8 = \frac{64}{3} = 21.3 km^{2}$

In the exam: in a multi-part problem, each result earns marks. Write the gradient before the equation, state what you substitute for each intercept with an axis (for example “ $y = 0$ ”) and set up the system before solving it, even if you use the GDC. Write the unit with the area.

Common mistakes

Summary

The gradient of the line through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ . Positive: rises; negative: falls; zero: horizontal; if $x_{1} = x_{2}$ the line is vertical and has no gradient.
With $θ$ the acute angle with the x-axis: $m = tan θ$ .
Point-gradient form: $y - y_{1} = m (x - x_{1})$ . Explicit form: $y = m x + c$ . General form: $a x + b y + d = 0$ , with $m = - \frac{a}{b}$ .
Horizontal lines $y = k$ and vertical lines $x = k$ .
Intercept with the x-axis: set $y = 0$ . Intercept with the y-axis: set $x = 0$ .
The intersection of two lines is the solution of the system formed by their equations; on the GDC, use the equation solver or the graph.
When interpreting a linear model $y = m x + c$ : $m$ is the change per unit and $c$ the initial value.

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