Math Functions Important types of functions and their properties Linear functions - Lines Exercises: Finding line equations

Exercises: Finding line equations

Set the equation of the straight line with gradient $m = - \frac{4}{3}$ through the point $P (- 2 | - 0.5)$ and draw it in a coordinate system.

For this task you need the following basic knowledge: Linear Function

Setting up line equations

$m = - \frac{4}{3}$ ; $P (- 2 | - 0.5)$

Plug $m$ and $P$ into the general linear equation and solve for $t$ .

$- \frac{1}{2}$	$=$	$- \frac{4}{3} \cdot (- 2) + t$
	↓	$- \frac{4}{3} \cdot (- 2) = \frac{(- 4) \cdot (- 2)}{3} = \frac{8}{3}$
$- \frac{1}{2}$	$=$	$\frac{8}{3} + t$	$- \frac{8}{3}$
$t$	$=$	$- \frac{1}{2} - \frac{8}{3}$
	↓	Reduce the two fractions to the same denominator.
$t$	$=$	$- \frac{3}{6} - \frac{16}{6}$
	↓	Subtract.
$t$	$=$	$- \frac{19}{6}$
	↓	Convert to a mixed fraction.
$t$	$=$	$- 3 \frac{1}{6}$
	↓	Setze t und m in die allgemeine Geradengleichung ein.

$\Rightarrow$ $y = - \frac{4}{3} x - 3 \frac{1}{6}$

Drawing the line

To determine $t$ , you substitute the $x$ - and $y$ -values of one of the two points into the equation of the line. Here, it is calculated exemplarily with the point $P$ ."

Choose any point on the line, for example, the given point $P (- 2 ∣ - 0.5)$ . Find a second point by moving from the point $P$ 1 unit to the right. According to the slope $m = - \frac{3}{4}$ , 1you then have to go $\frac{4}{3}$ units down. You obtain the green slope triangle. Connect the two points to form a line.

Alternatively, you can find the second point by moving 3 units to the right and 4 units down. You obtain the orange slope triangle. (The slope is still $m = - \frac{3}{4}$ .)

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