Exercises: Distances, parallel and perpendicular lines

Consider the equation $y = \frac{3}{2} x + 1$ .

Zeichne die Gerade zu der Gleichung in ein Koordinatensystem.

For this task you need the following basic knowledge: Lines in coordinate systems
Choose any point on the line, e.g. the $y$ -axis-intercept $(0 | 1)$ , and go two to the right and three upwards, which corresponds to a slope of $m = \frac{3}{2}$ . Connect the two points to obtain a straight line.
Set up the equation of the perpendicular line through the point $P (3 | 2.25)$ .

For this task you need the following basic knowledge: Slope/Gradient of a line
You obtain the slope of the perpendicular (=normal) line by dividing -1 by the original slope.
$m$ $=$ $- 1 : \frac{3}{2} = \frac{- 1}{1} \cdot \frac{2}{3} = - \frac{2}{3}$
↓
Plug the coordinates of $P$ and $m$ into the general line equation.
$2.25$ $=$ $- \frac{2}{3} \cdot 3 + t$
$=$ $- \frac{6}{3} + t$
$=$ $- \frac{2}{1} + t$
$=$ $- 2 + t$ $+ 2$
↓
Solve for $t$ .
$t$ $=$ $2.25 + 2 = 4.25$
Substitute $m$ and $t$ into the general line equation.
$y$ $=$ $- \frac{2}{3} x + 4.25$
Draw the line in the same coordinate system as the line from exercise 1.

For this task you need the following basic knowledge: Slope/Gradient of a line
Choose any point on the line, e.g. the given point $P (3 | 2.25)$ and go, according to the gradient $m = - \frac{2}{3}$ , $3$ units to the left and $2$ units upwards. Connect the two points to obtain a straight line. (see sketch in subtask 1)
Calculate the intersection of the two lines.

Calculating the $x$ -coordinate
Set the straight line functions equal to obtain the point of intersection.
$\frac{3}{2} x + 1$ $=$ $- \frac{2}{3} x + \frac{17}{4}$ $+ \frac{2}{3} x | - 1$
↓
Get $x$ to one side.
$\frac{3}{2} x + \frac{2}{3} x$ $=$ $\frac{17}{4} - 1$
↓
Factor out $x$ and add the fractions.
$\frac{13}{6} x$ $=$ $\frac{13}{4}$ $: \frac{13}{6}$
$x$ $=$ $\frac{6}{4} = \frac{3}{2} = 1.5$
Calculating the $y$ -coordinate

Plug $x$ into one of the line equations, e.g. that of the given line.
$y$ $=$ $\frac{3}{2} \cdot \frac{3}{2} + 1$
$=$ $\frac{9}{4} + 1$
$=$ $3.25$
$\Rightarrow$ S $(1.5 | 3.25)$

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$m$	$=$	$- 1 : \frac{3}{2} = \frac{- 1}{1} \cdot \frac{2}{3} = - \frac{2}{3}$
	↓	Plug the coordinates of $P$ and $m$ into the general line equation.
$2.25$	$=$	$- \frac{2}{3} \cdot 3 + t$
	$=$	$- \frac{6}{3} + t$
	$=$	$- \frac{2}{1} + t$
	$=$	$- 2 + t$	$+ 2$
	↓	Solve for $t$ .
$t$	$=$	$2.25 + 2 = 4.25$

$\frac{3}{2} x + 1$	$=$	$- \frac{2}{3} x + \frac{17}{4}$	$+ \frac{2}{3} x \| - 1$
	↓	Get $x$ to one side.
$\frac{3}{2} x + \frac{2}{3} x$	$=$	$\frac{17}{4} - 1$
	↓	Factor out $x$ and add the fractions.
$\frac{13}{6} x$	$=$	$\frac{13}{4}$	$: \frac{13}{6}$
$x$	$=$	$\frac{6}{4} = \frac{3}{2} = 1.5$

$y$	$=$	$\frac{3}{2} \cdot \frac{3}{2} + 1$
	$=$	$\frac{9}{4} + 1$
	$=$	$3.25$

Exercises: Distances, parallel and perpendicular lines

Calculating the x-coordinate

﻿Calculating the y-coordinate

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Calculating the $x$ -coordinate

Calculating the $y$ -coordinate