Exercises: Distances, parallel and perpendicular lines

Determine the equation of the straight line through ...

the point $P (- 3 | 4)$ and being parallel to the $x$ -axis.

For this task you need the following basic knowledge: Line equation
Parallel to the $x$ -axis, means that the slope is $m = 0$ .
Plug $m$ and $P$ into the general line equation.
$\begin{array}{rcl} 4 & = & 0 \cdot (- 3) + t \\ 4 & = & t \end{array}$
Assemble to a line equation.
$y = 0 \cdot x + 4 = 4$
the point $Q (2 | 5)$ and is parallel to the bisector of the 2nd quadrant (the diagonal pointing down).

For this task you need the following basic knowledge: Line equation
Parallel to the bisector of the 2nd quadrant means the line has the same slope as the bisector.
The slope of the bisector of the 2nd quadrant is -1
$\Rightarrow m = - 1$
Plug $m$ into the straight line equation in order to calculate $t$ .
$\begin{array}{rcl} 5 & = & - 2 + t & | + 2 \\ t & = & 7 \end{array}$
Plug $m$ and $t$ into the general equation of a straight line.
$y = - 1 \cdot x + 7 = - x + 7$
the point $R (- 4 | 2)$ and is parallel to the $y$ -axis.

For this task you need the following basic knowledge: Line equation
Parallel to the $y$ -axis means that there is no function equation, since an $x$ -value is assigned an infinite number of $y$ -values.
The straight line can therefore only be described as the $x$ -value of $R$ .
$\Rightarrow x = - 4$
the point $S (2 | - 3)$ and is parallel to the bisector of the 1st quadrant (the diagonal pointing up).

For this task you need the following basic knowledge: Line equation
The line being parallel to the bisector of the 1st quadrant means it has the same slope.
The slope of the bisector of the 1st quadrant is $1$ .
$\Rightarrow m = 1$
Substitute $m$ and $S$ into the general line equation and solve for $t$ .
$\begin{array}{rcl} - 3 & = & 2 + t & | - 2 \\ t & = & - 5 \end{array}$
Substitute $m$ and $t$ into the general line equation.
$\Rightarrow y = x - 5$
the origin and is parallel to the straight line $AB$ with $A (- 72 | - 60)$ and $B (- 24 | - 20)$ .

For this task you need the following basic knowledge: Line equation
Passing through the origin means, the $y$ -axis intercept is $t = 0$ .
Being parallel to the straight line $AB$ means that the line has the same slope as $AB$ .
$A (- 72 | - 60); B (- 24 | - 20)$
Calculate the slope $m$ using the difference quotient .
$m = \frac{- 20 - (- 60)}{- 24 - (- 72)} = \frac{40}{48} = \frac{5}{6}$
Substitute $m$ and $t$ into the general line equation.
$\Rightarrow y = \frac{5}{6} x$

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