The equation of a line from a point and a gradient, or from two points – Linear and Quadratic Functions (Senior 3 Mathematics) | Ganzaa

Linear and Quadratic Functions

Lessons

Senior 3 · Mathematics

Linear and Quadratic Functions

5 lessons

Lesson 2 of 5

The equation of a line from a point and a gradient, or from two points

25 min

Two readings, one line

Air gets colder as you climb. Suppose a weather station at a height of 1,500 m records $21\,^\circ\mathrm{C}$ and, at the same moment, a station at 2,500 m records $15\,^\circ\mathrm{C}$ (illustrative readings). If the temperature falls at a steady rate, the two readings fix a straight line, and that line lets you estimate the temperature at any height in between.

A line is fixed by two facts: either one point and the gradient, or two points. This lesson turns each into an equation.

Important

The point–gradient form. The line with gradient $m$ through the point $(x_{1}, y_{1})$ has equation

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

Why: take any other point $(x, y)$ on the line. The gradient between $(x_{1}, y_{1})$ and $(x, y)$ must be $m$ , so $\frac{y - y_1}{x - x_1} = m$ . Multiply both sides by $x - x_{1}$ .

Worked example 1: a point and a gradient

Find the equation of the line with gradient $- 2$ through $(3, 1)$ . Give it in the form $y = m x + c$ and in general form.

Solution

y - 1 = -2(x - 3) \;\Longrightarrow\; y - 1 = -2x + 6 \;\Longrightarrow\; y = -2x + 7.

General form: $2 x + y - 7 = 0$ . Check the point: $2 (3) + 1 - 7 = 0$ ✓.

Important

A line through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$

Find the gradient $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
Use the point–gradient form with either point.
Check that the other point satisfies your equation.

The same steps in one formula (the two-point form), when $x_1 \ne x_2$ and $y_1 \ne y_2$ :

\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}.

Worked example 2: two points

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

Solution Gradient: $m = \frac{-4 - 4}{3 - (-1)} = \frac{-8}{4} = -2$ . Use point $A$ :

y - 4 = -2\big(x - (-1)\big) = -2(x + 1) \;\Longrightarrow\; y = -2x + 2.

General form: $2 x + y - 2 = 0$ . Check $B$ : $2 (3) + (- 4) - 2 = 0$ ✓.

Across 4, down 8: the gradient is −2, and the line cuts the y-axis at (0, 2).

Worked example 3: the temperature model

Use the readings $(1500, 21)$ and $(2500, 15)$ , where $h$ is the height in metres and $T$ the temperature in $^\circ\mathrm{C}$ . (a) Find $T$ in terms of $h$ . (b) Estimate the temperature at 2,000 m. (c) At what height would it be $12\,^\circ\mathrm{C}$ ?

Solution (a) $m = \frac{15 - 21}{2500 - 1500} = \frac{-6}{1000} = -0.006$ : the temperature falls $0.006\,^\circ\mathrm{C}$ per metre, or $6\,^\circ\mathrm{C}$ per kilometre.

T - 21 = -0.006(h - 1500) \;\Longrightarrow\; T = -0.006h + 30.

(b) $T = -0.006(2000) + 30 = -12 + 30 = 18\,^\circ\mathrm{C}$ , halfway between the readings, as it should be.

(c) $12 = - 0.006 h + 30$ , so $0.006 h = 18$ and $h = 3000$ m. This is outside the measured range, so it is only an estimate: the line is a model, and real air does not cool at exactly the same rate everywhere.

Worked example 4: special lines and three points

(a) Find the line through $(2, 5)$ and $(2, - 1)$ , and the line through $(- 3, 4)$ and $(5, 4)$ . (b) Do $A (0, 1)$ , $B (2, 5)$ and $C (5, 11)$ lie on one line?

Solution (a) Both points of the first pair have $x = 2$ : the line is vertical, $x = 2$ . Both points of the second pair have $y = 4$ : the line is horizontal, $y = 4$ (gradient $\frac{0}{8} = 0$ ).

(b) $m_{AB} = \frac{5 - 1}{2 - 0} = 2$ and $m_{BC} = \frac{11 - 5}{5 - 2} = 2$ . The gradients are equal and the lines share the point $B$ , so $A$ , $B$ and $C$ are collinear (on one line), $y = 2 x + 1$ .

Definition

Collinear points: points that lie on one straight line.

Test: $A$ , $B$ and $C$ are collinear exactly when $m_{AB} = m_{BC}$ (or when all three share the same $x$ ).

Example: $(0, 1)$ , $(2, 5)$ and $(5, 11)$ are collinear; $(0, 1)$ , $(2, 5)$ and $(5, 12)$ are not, because $\frac{12 - 5}{3} = \frac{7}{3} \ne 2$ .

Warning

Watch the double minus. Through $(- 1, 4)$ , the form is $y - 4 = m\big(x - (-1)\big) = m(x + 1)$ , not $m (x - 1)$ .

A vertical line has no $y = m x + c$ form. Through $(2, 5)$ and $(2, - 1)$ the gradient formula divides by 0; write $x = 2$ instead.

Tip

Always check the second point. Substituting the point you did not use takes ten seconds and catches almost every sign error.

Khan Academy: Point-slope and slope-intercept form from two points. Watch how the gradient is found first, then the point–gradient form is written and rearranged to y = mx + c.

✦

Practice

Give each answer in the form $y = m x + c$ and in general form, unless the line is vertical. The last three are harder.

The line with gradient 3 through $(2, - 1)$ .
The line with gradient $-\frac{1}{2}$ through $(4, 3)$ .
The line through $(1, 2)$ and $(4, 11)$ .
The line through $(- 2, 3)$ and $(4, - 1)$ .
The line through $(5, - 2)$ and $(5, 6)$ .
Are the points $(- 1, - 5)$ , $(2, 1)$ and $(4, 5)$ collinear? Show why.
A line cuts the axes at $(6, 0)$ and $(0, 4)$ . Find its equation.
A phone battery shows 80% at 09:00 and 50% at 11:30. Assuming it drains at a steady rate, write the charge $B$ in terms of $t$ , the hours after 09:00, and find when it reaches 0%.

Show answersHide answers

$y + 1 = 3 (x - 2)$ : $y = 3 x - 7$ ; $3 x - y - 7 = 0$ .
$y - 3 = -\frac{1}{2}(x - 4)$ : $y = -\frac{1}{2}x + 5$ ; $x + 2 y - 10 = 0$ .
$m = \frac{9}{3} = 3$ : $y = 3 x - 1$ ; $3 x - y - 1 = 0$ .
$m = \frac{-4}{6} = -\frac{2}{3}$ : $y = -\frac{2}{3}x + \frac{5}{3}$ ; $2 x + 3 y - 5 = 0$ .
Vertical: $x = 5$ .
Gradients $\frac{6}{3} = 2$ and $\frac{4}{2} = 2$ are equal, with the middle point shared: yes, collinear.
$m = \frac{4 - 0}{0 - 6} = -\frac{2}{3}$ : $y = -\frac{2}{3}x + 4$ ; $2 x + 3 y - 12 = 0$ .
$m = \frac{50 - 80}{2.5} = -12$ per hour: $B = 80 - 12 t$ . $B = 0$ when $t = \frac{80}{12} = 6\frac{2}{3}$ h, that is, 6 h 40 min after 09:00: at 15:40.

Sources and credits

Written by Ganzaa.
Video: Point-slope and slope-intercept form from two points, Khan Academy, embedded from YouTube.

The graph, worked examples, practice items and quiz questions were written by Ganzaa; the temperature readings and battery figures are illustrative.

1Two readings, one line

Two readings, one line