Exercises: Distances, parallel and perpendicular lines

1
Determine the equation of the line that passes through the point $P$ and is perpendicular to the given line $y = m x + t$ .
1. $y = 3 x + 2$
  $P (3 | 5)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  First determine the slope of the line $h$ perpendicular to $g (x) : y = 3 x + 2$ , where being perpendicular implies
  $m_{2} = - \frac{1}{m_{1}}$
  Plug in the known values.
  $m_{2} = - \frac{1}{3}$
  So the line we are looking for has a slope of $- \frac{1}{3}$ .
  Now determine the $y$ -axis intercept of the line $h$ through $P (3 | 5)$ by substituting the point $P$ into the general line equation.
  $h (x_{p}) : y_{p} = m_{2} x_{p} + b$
  Plug in the values.
  $5 = - \frac{1}{3} \cdot 3 + b | + 1$
  Simplify and add $1$ .
  $6 = b \Leftrightarrow b = 6$
  So the line equation is $h (x) : y = - \frac{1}{3} x + 6$ .
  Do you have a question?
  Write it here 👉
2. $y = 0.5 x + 1$
  $P (1 | 2)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  First determine the slope of the line $h$ perpendicular to $g (x) : y = 0.5 x + 1$ , where being perpendicular implies
  $m_{2} = - \frac{1}{m_{1}}$
  Plug in the known values.
  $m_{2} = - \frac{1}{0.5} = - \frac{1}{\frac{1}{2}} = - 2$
  So the line we are looking for has a slope of $- 2$ .
  Now determine the $y$ -axis intercept of the line $h$ through $P (1 | 2)$ by substituting the point $P$ into the general line equation.
  $h (x_{p}) : y_{p} = m_{2} x_{p} + b$
  Plug in the values.
  $2 = - 2 \cdot 1 + b$ $| + 2$
  Simplify and add 2.
  $4 = b \Rightarrow b = 4$
  So the line equation is $h (x) : y = - 2 x + 4$ .
  Do you have a question?
  Write it here 👉
3. $y = - 5 x + 6$
  $P (- 10 | 1)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  First determine the slope of the line $h$ perpendicular to $g (x) : y = - 5 x + 6$ , where being perpendicular implies
  $m_{2} = - \frac{1}{m_{1}}$
  Plug in the known values.
  $m_{2} = - \frac{1}{- 5} = 0.2$
  So the line we are looking for has a slope of $0.2$ .
  Now determine the $y$ -axis intercept of the line $h$ through $P (- 10 | 1)$ by substituting the point $P$ into the general line equation.
  $h (x_{p}) : y_{p} = m_{2} x_{p} + b$
  Plug in the values.
  $1 = 0.2 \cdot (- 10) + b$ $| + 2$
  Simplify and add 2.
  $3 = b \Rightarrow b = 3$
  So the line equation is $h (x) : y = 0.2 x + 3$ .
  Do you have a question?
  Write it here 👉
4. $y = 4 x + 3$
  $P (2 | - 5)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  First determine the slope of the line $h$ perpendicular to $g (x) : y = 4 x + 3$ , where being perpendicular implies
  $m_{2} = - \frac{1}{m_{1}}$
  Plug in the known values.
  $m_{2} = - \frac{1}{4} = - 0.25$
  So the line we are looking for has a slope of $- 0.25$ .
  Now determine the $y$ -axis intercept of the line $h$ through $P (2 | - 5)$ by substituting the point $P$ into the general line equation.
  $h (x_{p}) : y_{p} = m_{2} x_{p} + b$
  Plug in the values.
  $- 5 = - 0.25 \cdot 2 + b$ $+ 0.5$
  Simplify and add $0.5$ .
  $- 4.5 = b \Rightarrow b = - 4.5$
  So the line equation is $h (x) : y = - 0.25 x - 4.5$ .
  Do you have a question?
  Write it here 👉
5. $y = - \frac{2}{3} x + 2$
  $P (4 | 6)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  First determine the slope of the line $h$ perpendicular to $g (x) : y = - \frac{2}{3} x + 3$ , where being perpendicular implies
  $m_{2} = - \frac{1}{m_{1}}$
  Plug in the known values.
  $m_{2} = - \frac{1}{- \frac{2}{3}} = \frac{3}{2}$
  So the line we are looking for has a slope of $\frac{3}{2}$ .
  Now determine the $y$ -axis intercept of the line $h$ through $P (4 | 6)$ by substituting the point $P$ into the general line equation.
  $h (x_{p}) : y_{p} = m_{2} x_{p} + b$
  Plug in the values.
  $6 = \frac{3}{2} \cdot 4 + b$ $| - 6$
  Simplify and subtract 6.
  $0 = b \Rightarrow b = 0$
  So the line equation is $h (x) : y = \frac{3}{2} x$ .
  Do you have a question?
  Write it here 👉
6. $y = \frac{1}{3} x - 2$
  $P (2 | 5)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  First determine the slope of the line $h$ perpendicular to $g (x) : y = \frac{1}{3} x - 2$ , where being perpendicular implies
  $m_{2} = - \frac{1}{m_{1}}$
  Plug in the known values.
  $m_{2} = - \frac{1}{\frac{1}{3}} = - 3$
  So the line we are looking for has a slope of $- 3$ .
  Now determine the $y$ -axis intercept of the line $h$ through $P (2 | 5)$ by substituting the point $P$ into the general line equation.
  $h (x_{p}) : y_{p} = m_{2} x_{p} + b$
  Plug in the values.
  $5 = - 3 \cdot 2 + b$ $| + 6$
  Simplify and add 6.
  $11 = b \Rightarrow b = 11$
  So the line equation is $h (x) : y = - 3 x + 11$ .
  Do you have a question?
  Write it here 👉
2
Determine the equation of the line $g$ that is parallel to the line $h$ and passes through the point $P$ .
1. $h$ : $y = 3 x - 2$ ; $P (1 | 0) $
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  $y = 3 x - 2$ ; $P (1 | 0)$
  For the line to be parallel to $h$ , it must have the same slope.
  The slope of a line is the variable $m$ of the general line equation.
  $m = 3$
  Setting up the line equation
  Plug $m = 3$ and $P (1 | 0)$ into the the general line equation.
  $0$ $=$ $3 \cdot 1 + t$ $- 3$
  ↓
  solve for $t$
  $t$ $=$ $- 3$
  Plug $m$ and $t$ into the general line equation.
  $\Rightarrow$ Line equation: $y = 3 x - 3$
  Do you have a question?
  Write it here 👉
2. $h$ : $y = x - 4$ ; $P (1 | 2) $
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  $y = x - 4$ ; $P (1 | 2)$
  For the line to be parallel to $h$ , it must have the same slope.
  The slope of a line is the variable $m$ of the general line equation.
  $m = 1$
  Setting up the line equation
  Plug $m = 1$ and $P (1 | 2)$ into the the general line equation.
  $2$ $=$ $1 + t$ $- 1$
  ↓
  solve for $t$
  $t$ $=$ $1$
  Plug $m$ and $t$ into the general line equation.
  $\Rightarrow$ Line equation: $y = x + 1$
  Do you have a question?
  Write it here 👉
3. $h$ : $y = 4 x$ ; $P (5 | 18) $
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  $y = 4 x$ ; $P (5 | 18)$
  For the line to be parallel to $h$ , it must have the same slope.
  The slope of a line is the variable $m$ of the general line equation.
  $m = 4$
  Setting up the line equation
  Plug $m = 4$ and $P (5 | 18)$ into the the general line equation.
  $18$ $=$ $4 \cdot 5 + t$ $- 20$
  ↓
  solve for $t$
  $t$ $=$ $- 2$
  Plug $m$ and $t$ into the general line equation.
  $\Rightarrow$ Line equation: $y = 4 x - 2$
  Do you have a question?
  Write it here 👉
4. $h$ : $y = - 2 x + 1$ ; $P (- 1 | 4)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  $y = - 2 x + 1$ ; $P (- 1 | 4)$
  For the line to be parallel to $h$ , it must have the same slope.
  The slope of a line is the variable $m$ of the general line equation.
  $m = - 2$
  Setting up the line equation
  Plug $m = - 2$ and $P (- 1 | 4)$ into the the general line equation.
  $4$ $=$ $- 2 \cdot (- 1) + t$ $- 2$
  ↓
  solve for $t$
  $t$ $=$ $2$
  Plug $m$ and $t$ into the general line equation.
  $\Rightarrow$ Line equation: $y = - 2 x + 2$
  Do you have a question?
  Write it here 👉
3
Determine the equation of the straight line through ...
1. the point $P (- 3 | 4)$ and being parallel to the $x$ -axis.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.
  $y = 0 \cdot x + 4 = 4$
  Do you have a question?
  Write it here 👉
2. the point $Q (2 | 5)$ and is parallel to the bisector of the 2nd quadrant (the diagonal pointing down).
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.
  $y = - 1 \cdot x + 7 = - x + 7$
  Do you have a question?
  Write it here 👉
3. the point $R (- 4 | 2)$ and is parallel to the $y$ -axis.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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 $R$ .
  $\Rightarrow x = - 4$
  Do you have a question?
  Write it here 👉
4. the point $S (2 | - 3)$ and is parallel to the bisector of the 1st quadrant (the diagonal pointing up).
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.
  $\Rightarrow y = x - 5$
  Do you have a question?
  Write it here 👉
5. the origin and is parallel to the straight line $AB$ with $A (- 72 | - 60)$ and $B (- 24 | - 20)$ .
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.
  $\Rightarrow y = \frac{5}{6} x$
  Do you have a question?
  Write it here 👉
4
Two perpendicular lines intersect at $S (- 2 | - 1)$ .
Determine at least one possible line equations.

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
We need to construct two perpendicular lines with the point of intersection $(- 2, - 1)$ . As you can already see from the problem setting, there are several ways to choose two such straight lines. One possibility is given below. A good criterion to check whether the two chosen straight lines are perpendicular to each other is to check whether $m_{1} \cdot m_{2} = - 1$ .
For instance, choose the line $g$ with $y = g (x) = x + 1$ and $h$ with $y = h (x) = - x - 3$ . Then $m_{1} \cdot m_{2} = 1 \cdot (- 1) = - 1$ and we have $g (- 2) = - 1$ and $h (- 2) = - 1$ . So the point of intersection is on the two lines and they are perpendicular to each other.
Caution:
You may also choose the line $y = - 1$ , which is parallel to the $x$ -axis and passes through the point $(- 2, - 1)$ . Then there is exactly one line perpendicular to it and passing through this point, namely that one given by $x = - 2$ and running parallel to the $y$ - axis. However, $x = - 2$ does not describe a function but a relation!
5
Calculate the distance of the line to the origin.
1. $y = \frac{3}{4} x - 5$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  The shortest connection of any point on the line $g (x) : y = m_{1} \cdot x + b$ to the origin is a second line $h (x) : y = m_{2} \cdot x + b_{2}$ which passes through the origin and is perpendicular to $g$ .
  Passing through the origin means $h (0) = b_{2} = 0$ .
  Being perpendicular means
  $m_{2} = - \frac{1}{m_{1}}$
  Plug in the known value of $m_{1}$ .
  $m_{2} = - \frac{1}{\frac{3}{4}} = - \frac{4}{3}$
  So the perpendicular line has the line equation $h (x) : y = - \frac{4}{3} x$ .
  Now calculate the intersection point $A (x_{s} | y_{s})$ of the two lines by setting equal their line equations.
  $g (x_{s})$ $=$ $h (x_{s})$
  ↓
  Plug in the line equations.
  $- \frac{4}{3} x_{s}$ $=$ $\frac{3}{4} x_{s} - 5$ $- \frac{3}{4} x_{s}$
  ↓
  Get the variable $x_{s}$ on the left side
  $- \frac{25}{12} x_{s}$ $=$ $- 5$ $\cdot (- 1)$
  $\frac{25}{12} x_{s}$ $=$ $5$ $: \frac{25}{12}$
  $x_{s}$ $=$ $\frac{12}{5}$
  $=$ $2.4$
  Now substitute $x_{s}$ into one of the line equations to determine $y_{s}$ .
  $h (x_{s}) : y_{s} = - \frac{4}{3} x_{s}$
  Plug in $x_{s} = 2.4$ .
  $y_{s} = - \frac{4}{3} \cdot 2.4 = - 3.2$
  The intersection of the lines $h$ and $g$ is therefore at $A (2.4 | - 3.2)$ .
  Now determine the distance of the origin to the calculated intersection point $A$ , this is exactly the shortest distance of the line $g$ to the origin.
  $d = \sqrt{(x_{s} - x_{0})^{2} + (y_{s} - y_{0})^{2}}$
  Plug in the values.
  $d = \sqrt{(2.4 - 0)^{2} + (- 3.2 - 0)^{2}}$
  Simplify.
  $= \sqrt{5.76 + 10.24} = \sqrt{16} = 4$
  The shortest distance of the line $g$ to the origin is therefore $4$ .
  Do you have a question?
  Write it here 👉
2. $y = - \frac{1}{2} x + 2$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  The shortest connection of any point on the line $g (x) : y = m_{1} \cdot x + b$ to the origin is a second line $h (x) : y = m_{2} \cdot x + b_{2}$ which passes through the origin and is perpendicular to $g$ .
  Passing through the origin means $h (0) = b_{2} = 0$ .
  Being perpendicular means
  $m_{2} = - \frac{1}{m_{1}}$
  Plug in the known value of $m_{1}$ .
  $m_{2} = - \frac{1}{- \frac{1}{2}} = 2$
  So the perpendicular line has the line equation $h (x) : y = 2 x$ .
  Now calculate the intersection point $A (x_{s} | y_{s})$ of the two lines by setting equal their line equations.
  $g (x_{s})$ $=$ $h (x_{s})$
  ↓
  Plug in the line equations.
  $- \frac{1}{2} x_{s} + 2$ $=$ $2 x_{s}$ $- 2 x_{s}$
  ↓
  Get the variable $x_{s}$ on the left side
  $- 2.5 x_{s} + 2$ $=$ $0$ $- 2$
  ↓
  Get the 2 to the right side.
  $- 2.5 x_{s}$ $=$ $- 2$ $: (- 2.5)$
  $x_{s}$ $=$ $0.8$
  Now substitute $x_{s}$ into one of the line equations to determine $y_{s}$ .
  $h (x_{s}) : y_{s} = 2 x_{s}$
  Plug in $x_{s} = 0.8$ .
  $y_{s} = 2 \cdot 0.8 = 1.6$
  The intersection of the lines $h$ and $g$ is therefore at $A (0.8 | 1.6)$ .
  Now determine the distance of the origin to the calculated intersection point $A$ , this is exactly the shortest distance of the line $g$ to the origin.
  $d = \sqrt{(x_{s} - x_{0})^{2} + (y_{s} - y_{0})^{2}}$
  Plug in the values.
  $d = \sqrt{(0.8 - 0)^{2} + (1.6 - 0)^{2}}$
  Simplify.
  $= \sqrt{0.64 + 2.56} = \sqrt{3.2} \approx 1.79$
  The shortest distance of the line $g$ to the origin is therefore approximately $1.79$ .
  Do you have a question?
  Write it here 👉
6
Calculate the distance of the parallel lines g: $y = - \frac{1}{2} x + 2$ and h: $y = - \frac{1}{2} x - 3$ .

Distance between two parallel lines
The shortest path between two parallel lines can be read off along a normal (=perpendicular line) of the two parallel lines .
The distance of the straight lines is equal to the distance between the two intersection points of the normal line with the parallel lines .
Sketch
The construction looks as follows:
Line equation
Establish the line equation of the normal line d.
First calculate the slope.
$m_{d} \cdot m_{g}$ $=$ $- 1$ $: m_{g}$
$m_{d}$ $=$ $\frac{- 1}{m_{g}}$
↓
Plug in the slope $m_{g} = - \frac{1}{2}$
$m_{d}$ $=$ $\frac{- 1}{- \frac{1}{2}}$
↓
Simplify the right-hand side
$m_{d}$ $=$ $- 2$
Now set up the line equation.
For this, you need a point $T$ , which lies on line h. For instance, you may take the $y$ -axis intercept of $h (x)$ , which is $T (0 | - 3)$ . But any other point is also correct.
Plug $T (0 | - 3)$ and $m_{d}$ into the general line equation for d to get the parameter $t$ .
$- 3$ $=$ $2 \cdot (0) + t$
↓
multiply out
$t$ $=$ $- 3$
↓
Plug $t$ and $m_{d}$ into the general line equation.
$y$ $=$ $2 x - 3$
Determine the intersection of d and g
First calculate the $x$ -coordinate of the intersection point $S$ .
Normal line d: $y = 2 x - 3$
Paralle line g: $y = - \frac{1}{2} x + 2$
Equate both function equations:
$2 x - 3$ $=$ $- \frac{1}{2} x + 2$ $+ 3$
$2 x$ $=$ $- \frac{1}{2} x + 5$ $+ \frac{1}{2} x$
$\frac{5}{2} x$ $=$ $5$ $: \frac{5}{2}$
$x$ $=$ $2$
Now calculate the $y$ -coordinate of the intersection point $S$ .
$y$ $=$ $2 x - 3$
↓
Plug in x = 2 .
$y$ $=$ $2 \cdot 2 - 3$
↓
compute
$y$ $=$ $1$
Hence the point $S$ has the following coordinates:
$\Rightarrow$ $S (2 | 1)$
Determine the distance between points S and T.
Determine the distance in $x$ -direction.
$T (0 | - 3)$ , $S (2 | 1)$
Calculate the difference of the $x$ -values of $S$ and $T$ .
$Δ x = 2 - 0 = 2$
Determine the distance in $y$ -direction.
$T (0 | - 3)$ , $S (2 | 1)$
Calculate the difference of the $y$ -values.
$Δ y = 1 - (- 3) = 4$
Determine the distance in a direct line between the points.
$Δ x = 2$
$Δ y = 4$
This is done by the Pythagoraen Theorem:
$d^{2}$ $=$ $2^{2} + 4^{2}$
↓
compute both squares
$d^{2}$ $=$ $4 + 16$
↓
add
$d^{2}$ $=$ $20$
↓
take the root
$d^{}$ $=$ $\sqrt{20}$
If you enter $\sqrt{20}$ into a calculator and round the result to two decimal places, you get:
$d = \sqrt{20} \approx 4.47$
Result
The distance between the two lines is $d = \sqrt{20} \approx 4.47$ .
7
Consider the equation $y = \frac{3}{2} x + 1$ .
1. Zeichne die Gerade zu der Gleichung in ein Koordinatensystem.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.
  Do you have a question?
  Write it here 👉
2. Set up the equation of the perpendicular line through the point $P (3 | 2.25)$ .
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.
  $y$ $=$ $- \frac{2}{3} x + 4.25$
  Do you have a question?
  Write it here 👉
3. Draw the line in the same coordinate system as the line from exercise 1.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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)
  Do you have a question?
  Write it here 👉
4. 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 $(1.5 | 3.25)$
  Do you have a question?
  Write it here 👉

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$g (x_{s})$	$=$	$h (x_{s})$
	↓	Plug in the line equations.
$- \frac{4}{3} x_{s}$	$=$	$\frac{3}{4} x_{s} - 5$	$- \frac{3}{4} x_{s}$
	↓	Get the variable $x_{s}$ on the left side
$- \frac{25}{12} x_{s}$	$=$	$- 5$	$\cdot (- 1)$
$\frac{25}{12} x_{s}$	$=$	$5$	$: \frac{25}{12}$
$x_{s}$	$=$	$\frac{12}{5}$
	$=$	$2.4$

$m_{d} \cdot m_{g}$	$=$	$- 1$	$: m_{g}$
$m_{d}$	$=$	$\frac{- 1}{m_{g}}$
	↓	Plug in the slope $m_{g} = - \frac{1}{2}$
$m_{d}$	$=$	$\frac{- 1}{- \frac{1}{2}}$
	↓	Simplify the right-hand side
$m_{d}$	$=$	$- 2$

$- 3$	$=$	$2 \cdot (0) + t$
	↓	multiply out
$t$	$=$	$- 3$
	↓	Plug $t$ and $m_{d}$ into the general line equation.
$y$	$=$	$2 x - 3$

$2 x - 3$	$=$	$- \frac{1}{2} x + 2$	$+ 3$
$2 x$	$=$	$- \frac{1}{2} x + 5$	$+ \frac{1}{2} x$
$\frac{5}{2} x$	$=$	$5$	$: \frac{5}{2}$
$x$	$=$	$2$

$y$	$=$	$2 x - 3$
	↓	Plug in x = 2 .
$y$	$=$	$2 \cdot 2 - 3$
	↓	compute
$y$	$=$	$1$

$d^{2}$	$=$	$2^{2} + 4^{2}$
	↓	compute both squares
$d^{2}$	$=$	$4 + 16$
	↓	add
$d^{2}$	$=$	$20$
	↓	take the root
$d^{}$	$=$	$\sqrt{20}$

$m$	$=$	$- 1 : \frac{3}{2} = \frac{- 1}{1} \cdot \frac{2}{3} = - \frac{2}{3}$
	↓	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

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Setting up the line equation

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Setting up the line equation

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Setting up the line equation

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Setting up the line equation

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Distance between two parallel lines

Sketch

Line equation

Determine the intersection of d and g

Determine the distance between points S and T.

Result

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Calculating the x-coordinate

﻿Calculating the y-coordinate

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Calculating the $x$ -coordinate

Calculating the $y$ -coordinate