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
  $\displaystyle m_2=-\frac1{m_1}$
  Plug in the known values.
  $\displaystyle m_2=-\frac1{3}$
  So the line we are looking for has a slope of $\displaystyle-\frac13$ .
  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.
  $5=-\frac13\cdot3+b\;\;\;\left|+1\right.$
  Simplify and add $1$ .
  $6=b\Leftrightarrow b=6$
  So the line equation is $\displaystyle h(x):y=-\frac13x+6$ .
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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
  $m_2=\displaystyle -\frac1{m_1}$
  Plug in the known values.
  $\displaystyle m_2=-\frac1{0.5}=-\frac1{\frac12}=-2$
  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.
  $2=-2\cdot1+b$ $∣ + 2$
  Simplify and add 2.
  $4=b\Rightarrow b=4$
  So the line equation is $h (x) : y = - 2 x + 4$ .
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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
  $\displaystyle m_2=-\frac1{m_1}$
  Plug in the known values.
  $\displaystyle m_2=-\frac1{-5}=0.2$
  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 $h (x) : y = 0.2 x + 3$ .
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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
  $\displaystyle m_2=-\frac1{m_1}$
  Plug in the known values.
  $m_2=-\displaystyle\frac14=-0.25$
  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.
  $-5=-0.25\cdot2+b$ $+ 0.5$
  Simplify and add $0.5$ .
  $-4.5=b\Rightarrow b=-4.5$
  So the line equation is $h (x) : y = - 0.25 x - 4.5$ .
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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
  $\displaystyle m_2=-\frac1{m_1}$
  Plug in the known values.
  $\displaystyle m_2=-\frac1{-\frac23}=\frac32$
  So the line we are looking for has a slope of $\displaystyle\frac32$ .
  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.
  $\displaystyle6=\frac32\cdot4+b$ $∣ - 6$
  Simplify and subtract 6.
  $0=b\Rightarrow b=0$
  So the line equation is $\displaystyle h(x):y=\frac32x$ .
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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 $\displaystyle g(x):y=\frac13x-2$ , where being perpendicular implies
  $m_2=\displaystyle-\frac1{m_1}$
  Plug in the known values.
  $m_2=\displaystyle -\frac1{\frac13}=-3$
  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.
  $5=-3\cdot2+b$ $∣ + 6$
  Simplify and add 6.
  $11=b\Rightarrow b=11$
  So the line equation is $h (x) : y = - 3 x + 11$ .
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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.
  $\displaystyle 0$ $=$ $\displaystyle 3\cdot1+t$ $\displaystyle -3$
  ↓
  solve for $t$
  $\displaystyle t$ $=$ $\displaystyle -3$
  Plug $m$ and $t$ into the general line equation.
  $\;\;\Rightarrow\;\;$ Line equation: $y = 3 x - 3$
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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.
  $\displaystyle 2$ $=$ $\displaystyle 1+t$ $\displaystyle -1$
  ↓
  solve for $t$
  $\displaystyle t$ $=$ $\displaystyle 1$
  Plug $m$ and $t$ into the general line equation.
  $\Rightarrow\$ Line equation: $y = x + 1$
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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.
  $\displaystyle 18$ $=$ $\displaystyle 4\cdot5+t$ $\displaystyle -20$
  ↓
  solve for $t$
  $\displaystyle t$ $=$ $\displaystyle -2$
  Plug $m$ and $t$ into the general line equation.
  $\Rightarrow$ Line equation: $y = 4 x - 2$
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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.
  $\displaystyle 4$ $=$ $\displaystyle −2⋅(−1)+t$ $\displaystyle -2$
  ↓
  solve for $t$
  $\displaystyle t$ $=$ $\displaystyle 2$
  Plug $m$ and $t$ into the general line equation.
  $\Rightarrow$ Line equation: $y = - 2 x + 2$
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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.
  $\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} 4&=&0\cdot(-3)+t \\ 4&=& t \end{array}$
  Assemble to a line equation.
  $\displaystyle y=0\cdot x+4=4$
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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
  $\displaystyle \Rightarrow m=-1$
  Plug $m$ into the straight line equation in order to calculate $t$ .
  $\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} 5&=&-2+t &|+2\\ t&=&7 \end{array}$
  Plug $m$ and $t$ into the general equation of a straight line.
  $\displaystyle y=-1\cdot x+7=-x+7$
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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 $R$ .
  $\displaystyle \;\;\Rightarrow\;\;x=-4$
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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 $1$ .
  $\displaystyle \Rightarrow m=1$
  Substitute $m$ and $S$ into the general line equation and solve for $t$ .
  $\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} -3&=&2+t &|-2\\ t&=&-5 \end{array}$
  Substitute $m$ and $t$ into the general line equation.
  $\displaystyle \;\Rightarrow\;\;y=x-5$
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5. the origin and is parallel to the straight line $\overline{\mathrm{AB}}$ 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 $\overline{\mathrm{AB}}$ means that the line has the same slope as $\overline{\mathrm{AB}}$ .
  $\displaystyle A(-72|-60);B(-24|-20)$
  Calculate the slope $m$ using the difference quotient .
  $\displaystyle m=\frac{-20-(-60)}{-24-(-72)}=\frac{40}{48}=\frac{5}{6}$
  Substitute $m$ and $t$ into the general line equation.
  $\displaystyle \Rightarrow y=\frac{5}{6}x$
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4
Two perpendicular lines intersect at $S\left(-2|-1\right)$ .
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!

Calculate the distance of the line to the origin.

$y=\frac{3}{4}x-5$

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

$\displaystyle m_2=-\frac1{m_1}$

Plug in the known value of $m_{1}$ .

$\displaystyle m_2=-\frac{1}{\frac34}=-\frac43$

So the perpendicular line has the line equation $\displaystyle h(x):y=-\frac43 x$ .

Now calculate the intersection point $A \,(x_s|y_s)$ of the two lines by setting equal their line equations.

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

Now substitute $x_{s}$ into one of the line equations to determine $y_{s}$ .

$\displaystyle h(x_s):y_s=-\frac43x_s$

Plug in $x_{s} = 2.4$ .

$\displaystyle y_s=-\frac43\cdot2.4=-3.2$

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.

$\displaystyle d=\sqrt{(x_s-x_0)^2+(y_s-y_0)^2}$

Plug in the values.

$\displaystyle d=\sqrt{(2.4-0)^2+(-3.2-0)^2}$

Simplify.

$\displaystyle =\sqrt{5.76+10.24}=\sqrt{16}=4$

The shortest distance of the line $g$ to the origin is therefore $4$ .

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$y=-\frac{1}{2}x+2$

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

$\displaystyle m_2=-\frac1{m_1}$

Plug in the known value of $m_{1}$ .

$\displaystyle m_2=-\frac{1}{-\frac12}=2$

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.

$\displaystyle g\left(x_s\right)$	$=$	$\displaystyle h\left(x_s\right)$
	↓	Plug in the line equations.
$\displaystyle \displaystyle -\frac12x_s+2$	$=$	$\displaystyle 2x_s$	$\displaystyle -2x_s$
	↓	Get the variable $x_{s}$ on the left side
$\displaystyle -2.5x_s+2$	$=$	$\displaystyle 0$	$\displaystyle -2$
	↓	Get the 2 to the right side.
$\displaystyle -2.5x_s$	$=$	$\displaystyle -2$	$\displaystyle :\left(-2.5\right)$
$\displaystyle x_s$	$=$	$\displaystyle 0.8$

Now substitute $x_{s}$ into one of the line equations to determine $y_{s}$ .

$h (x_{s}) : y_{s} = 2 x_{s}$

Plug in $x_{s} = 0.8$ .

$y_s=2\cdot0.8=1.6$

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.

$\displaystyle d=\sqrt{(x_s-x_0)^2+(y_s-y_0)^2}$

Plug in the values.

$d=\sqrt{(0.8-0)^2+(1.6-0)^2}$

Simplify.

$=\sqrt{0.64+2.56}=\sqrt{3.2}\approx1.79$

The shortest distance of the line $g$ to the origin is therefore approximately $1.79$ .

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Calculate the distance of the parallel lines g: $y=-\frac{1}{2}x+2$ and h: $y=-\frac12x-3$ .

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:

Graph Geraden Parallelen Abstand zwischen den Parallelen

Line equation

Establish the line equation of the normal line d.

First calculate the slope.

$\displaystyle m_d\cdot m_g$	$=$	$\displaystyle -1$	$\displaystyle :m_g$
$\displaystyle m_d$	$=$	$\displaystyle \frac{-1}{m_g}$
	↓	Plug in the slope $m_g = -\frac{1}{2}$
$\displaystyle m_d$	$=$	$\displaystyle \frac{-1}{-\frac{1}{2}}$
	↓	Simplify the right-hand side
$\displaystyle m_d$	$=$	$\displaystyle -2$

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 $h\left(x\right)$ , which is $T \left(0\vert-3\right)$ . But any other point is also correct.

Plug $T\left(0\vert -3\right)$ and $m_{d}$ into the general line equation for d to get the parameter $t$ .

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

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: $y=-\frac12x+2$

Equate both function equations:

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

Now calculate the $y$ -coordinate of the intersection point $S$ .

$\displaystyle y$	$=$	$\displaystyle 2x-3$
	↓	Plug in x = 2 .
$\displaystyle y$	$=$	$\displaystyle 2\cdot2-3$
	↓	compute
$\displaystyle y$	$=$	$\displaystyle 1$

Hence the point $S$ has the following coordinates:

$\Rightarrow$ $S\left(2|1\right)$

Determine the distance between points S and T.

Determine the distance in $x$ -direction.

$T\left(0|-3\right)$ , $S\left(2|1\right)$

Calculate the difference of the $x$ -values of $S$ and $T$ .

$\Delta x=2-0=2$

Determine the distance in $y$ -direction.

$T\left(0|-3\right)$ , $S\left(2|1\right)$

Calculate the difference of the $y$ -values.

$\Delta y=1-(-3)=4$

Determine the distance in a direct line between the points.

$\Delta x=2$

$\Delta y=4$

This is done by the Pythagoraen Theorem:

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

If you enter $\sqrt{20}$ into a calculator and round the result to two decimal places, you get:

$d=\sqrt{20}\approx4.47$

Result

The distance between the two lines is $d=\sqrt{20}\approx4.47$ .

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

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 $(0\vert1)$ , and go two to the right and three upwards, which corresponds to a slope of $m=\frac32$ . Connect the two points to obtain a straight line.
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Set up the equation of the perpendicular line through the point $P (3 ∣ 2.25)$ .

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 $m=-\frac23$ , $3$ units to the left and $2$ units upwards. Connect the two points to obtain a straight line. (see sketch in subtask 1)
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Calculate the intersection of the two lines.

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$\displaystyle m$	$=$	$\displaystyle -1:\frac{3}{2}=\frac{-1}{1}\cdot\frac{2}{3}=-\frac{2}{3}$
	↓	Plug the coordinates of $P$ and $m$ into the general line equation.
$\displaystyle 2.25$	$=$	$\displaystyle -\frac{2}{3}\cdot3+t$
	$=$	$\displaystyle -\frac{6}{3}+t$
	$=$	$\displaystyle -\frac{2}{1}+t$
	$=$	$\displaystyle -2+t$	$\displaystyle +2$
	↓	Solve for $t$ .
$\displaystyle t$	$=$	$\displaystyle 2.25+2=4.25$

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

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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$\displaystyle 0$	$=$	$\displaystyle 3\cdot1+t$	$\displaystyle -3$
	↓	solve for $t$
$\displaystyle t$	$=$	$\displaystyle -3$

$\displaystyle 2$	$=$	$\displaystyle 1+t$	$\displaystyle -1$
	↓	solve for $t$
$\displaystyle t$	$=$	$\displaystyle 1$

$\displaystyle 18$	$=$	$\displaystyle 4\cdot5+t$	$\displaystyle -20$
	↓	solve for $t$
$\displaystyle t$	$=$	$\displaystyle -2$

$\displaystyle 4$	$=$	$\displaystyle −2⋅(−1)+t$	$\displaystyle -2$
	↓	solve for $t$
$\displaystyle t$	$=$	$\displaystyle 2$

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

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 xxx-coordinate

﻿Calculating the yyy-coordinate

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Calculating the $x$ -coordinate

Calculating the $y$ -coordinate