Exercises: Finding line equations

1
The following function graphs are given:
1. Which of the four graphs belongs to the equation $y=\frac{5}{4}x-1$ ?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equations
  You are given the graph equation: $y=\frac54x-1$
  You can read the slope and the y-intercept of this graph from the equation.
  $m=\frac{5}{4}$
  $ t = - 1$
  First check for which functions the y-intercept is $t = - 1$ by reading the y-value of each graph by cutting the y-axis.
  Only graph I and II have the y-intercept $- 1$ so you can exclude any other graph.
  Now check which of the two graphs has the slope $m=\frac{5}{4}$ by going one unit to the right from the point $x = 0$ and checking which of the two $y$ -values increases by $\frac{5}{4}$ .
  Both graphs start at the point $P (0; - 1)$ . As the straight line you are looking for has a gradient of $\frac{5}{4}$ , it also passes through the point $(0 + 4 ∣ - 1 + 5) = (4 ∣ 4)$ .
  Only line II runs through this point.
  $\Rightarrow$ Graph II is the one that belongs to the given equation.
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2. Wie lautet die Gleichung zum Graphen III?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equations
  We have to check graph III
  First read where the graph intersects the y-axis to determine the y-intercept.
  The y-value of the point where the y-axis is intersected is $y = 1.25$ . Therefore, $t = 1.25$ .
  Now read off how much the y-value changes if you go one to the right starting from $x = 0$ . This will give you the gradient.
  The y-value increases from $y = 1.25$ to $y = 2.25$ . The gradient is therefore $m=\frac{2{,}25-1{,}25}{1}=\frac11=1$ .
  Set up the equation.
  ⇒ Graph III has the equation $y = x + 1.25$ .
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  Write it here 👉
2
The following diagrams contain graphs of linear functions. Determine the function terms.
1. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear Functions
  Linear function $f (x)$
  Set up the general form of a linear function.
  $f(x)=m \cdot x+t$
  Read off the y-intercept from the diagram.
  $t = 2$
  Read off two points from the graph of the linear function to calculate the gradient:
  $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$
  You can use these points, for example:
  $P_{1} (- 3 ∣ 0)$ $x_{1} = - 3$ and $y_{1} = 0$
  $P_{2} (0 ∣ 2)$ $x_{2} = 0$ and $y_{2} = 2$
  For the gradient, you then get by substituting:
  $\displaystyle m = \frac{2 - 0}{0 - (-3)} = \frac{2}{3}$
  Now plug the calculated values of $m$ and $t$ into the general form:
  $f(x) = \dfrac{2}{3}\cdot x + 2$
  Linear function $g (x)$
  Set up the general form of a linear function.
  $g(x)=m \cdot x+t$
  Read off two points on the graph of the linear function to calculate the gradient.
  $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$
  For example, you can use these two points:
  $P_{1} (2 ∣ - 4)$ $x_{1} = 2$ and $y_{1} = - 4$
  $P_{2} (3.5 ∣ 0)$ $x_{2} = 3.5$ and $y_{2} = 0$
  The gradient is then:
  $\displaystyle m = \frac{0 - (-4)}{3.5 - 2} = \frac{4}{1.5} = 4 \cdot \frac{2}{3} = \frac{8}{3}$
  As the y-intercept is not visible, you have to calculate it. To do this, set up the equation of the line.
  $\displaystyle g(x) = \frac{8}{3} \cdot x + t$
  Insert one of the points, for example $(2 ∣ - 4)$ .
  $\displaystyle -4 = \frac{8}{3} \cdot 2 + t$
  Now solve for $t$ .
  $\displaystyle t = -4- \frac{8}{3} \cdot 2 = -\frac{12}{3}- \frac{16}{3} = -\frac{28}{3}$
  Substitute the values of $m$ and $t$ into the general form of the linear function, and you will get the linear equation:
  $\displaystyle g(x)=\frac{8}{3} \cdot x - \frac{28}{3}$
  Linear function $h (x)$
  Set up the general form of a linear function.
  $h(x)=m \cdot x+t$
  Read off two points from the graph of the linear function to calculate the gradient:
  $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$
  You can use these two points, for example:
  $P_{1} (0 ∣ - 2)$ $x_{1} = 0$ and $y_{1} = - 2$
  $P_{2} (4 ∣ - 3)$ $x_{2} = 4$ and $y_{2} = - 3$
  Now use them to calculate the gradient:
  $\displaystyle m = \frac{-3 - (-2)}{4 - 0} = \frac{-3 + 2}{4} = -\frac{1}{4}$
  Either read off $t = - 2$ or calculate the value. To calculate it, set up the linear equation.
  $\displaystyle h(x) = -\frac{1}{4} \cdot x + t$
  Insert a point that lies on the straight line, for example $(4 ∣ - 3)$ .
  $\displaystyle -3 = -\frac{1}{4} \cdot 4 + t$
  Now, solve for $t$ .
  $t = - 3 + 1 = - 2$
  Setze $m = -\dfrac14$ und $t = - 2$ in die allgemeine Form ein und du erhältst die Geradengleichung:
  $\displaystyle h(x) = -\frac{1}{4} \cdot x -2$
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  Write it here 👉
2. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear Functions
  Linear function $f (x)$
  Set up the general form of a linear function.
  $f(x)=m \cdot x+t$
  Read off two points from the graph of the linear function to calculate the gradient:
  $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$
  You can use these two points, for example:
  $P_{1} (2 ∣ 4)$ $x_{1} = 2$ and $y_{1} = 4$
  $P_2(2{,}5|0)$ $x_2= 2{,}5$ and $y_{2} = 0$
  The result is a gradient:
  $\displaystyle m = \frac{0 - 4}{2{,}5 - 2} = \frac{-4}{0{,}5} = -4 \cdot \frac{2}{1} = -8$
  As the y-intercept is not visible, you have to calculate it. Therefore, set up the equation of the line:
  $\displaystyle f(x) = -8 \cdot x + t$
  Insert one of the points, for example $(2 ∣ 4)$ :
  $\displaystyle 4 = -8 \cdot 2 + t$
  Solve for $t$ .
  $\displaystyle t = 4 + 8 \cdot 2 = 4 + 16 = 20$
  Plug $m = - 8$ and $t = 20$ into the general form, and you will get the linear equation as a result:
  $f(x) = -8\cdot x + 20$
  Linear function $g (x)$
  Set up the general form of a linear function.
  $g(x)=m \cdot x+t$
  Read off two points from the graph of the linear function to calculate the gradient:
  $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$
  You can use these two points, for example:
  $P_{1} (0 ∣ - 2)$ $x_{1} = 0$ and $y_{1} = - 2$
  $P_{2} (- 4 ∣ - 3)$ $x_{2} = - 4$ and $y_{2} = - 3$
  Now use them to calculate the gradient:
  $\displaystyle m = \frac{-3 - (-2)}{-4 - 0} = \frac{-3 +2}{-4} = \frac{1}{4}$
  Read off the y-intercept from the diagram.
  $t = - 2$
  Plug $m = \dfrac14$ and $t = - 2$ into the general form of the linear function, and you get the line equation of $g$ :
  $\displaystyle g(x)=\frac{1}{4} \cdot x - 2$
  Do you have a question?
  Write it here 👉
3
Determine the equation of the following straight line:

Für diese Aufgabe benötigst Du folgendes Grundwissen: Lines in Coordinate Systems
The general line equation is:
$\mathrm y=\mathrm m\cdot\mathrm x+\mathrm t$
Read the y-intercept $t$ , i.e. the point at which the straight line intersects the y-axis, from the drawing.
$t = - 1$
Find two points with (at best) integer coordinates.
$P\left(2 \;|\;2\right)$ and $Q(4\;| \;5)$ are on the line
There are two ways to determine the gradient $m$ :
1. $\;m=\displaystyle\frac{y_Q-y_P}{x_Q-x_P}$
Plug in the coordinates of $P$ and $Q$ .
$m=\displaystyle\frac{5-2}{4-2}=\frac32=1.5$
2.
Draw a slope triangle between the points. The vertical distance is the numerator, the horizontal distance is the denominator of the fraction that describes the gradient.
$m=\displaystyle\frac{\text{senkrecht}}{\text{waagerecht}}=\frac32=1.5$
The line equation is therefore given by :
$\mathrm g\left(\mathrm x\right)=\frac32\cdot\mathrm x-1=1.5x-1$

Calculate the gradient of the straight line through the given points.

$A (5 ∣ 7)$ , $B (- 3 ∣ 8)$

$A (1 ∣ 2)$ , $B (3 ∣ 4)$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a Line
$A (1 ∣ 2), B (3 ∣ 4)$
Insert the points into the difference quotient for the line.
$\displaystyle m$ $=$ $\displaystyle \frac{4-2}{3-1}$
↓
Simplify numerator and denominator
$\displaystyle m$ $=$ $\displaystyle \frac22$
$\displaystyle m$ $=$ $\displaystyle 1$
The gradient of the line is m = 1.
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Write it here 👉

Solve the following exercises.

What is the gradient of the straight line through the points $P (0 ∣ 3)$ and $Q (2 ∣ - 3)$ ?

Set up the equation of the line through the points $P (1 ∣ 3)$ and $Q (3 ∣ - 1)$ .

Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equation
You are given the points $P (1 ∣ 3)$ and $Q (3 ∣ - 1)$
Calculate the difference between the two x-values and the two y-values to determine the gradient.
x-values: $3 - 1 = 2$
y-values: $- 1 - 3 = - 4$
While the x-value increases by 2, the y-value decreases by 4. Divide the y-value by the x-value to calculate the gradient.
$m=\frac{-4}{2}=-2$
Substitute m, the x-value and the y-value of one of the points into the general line equation to determine $t$ .
$\displaystyle 3$ $=$ $\displaystyle -2\cdot1+t$
$\displaystyle 3$ $=$ $\displaystyle -2+t$ $\displaystyle +2$
$\displaystyle t$ $=$ $\displaystyle 5$
Plug $m$ and $t$ into the general linear equation to set up the equation of the function.
$y = - 2 x + 5$
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Write it here 👉

6
What is the gradient of the line through the points $\mathrm{P}\left(0|3\right)$ and $\mathrm{Q}\left(2|-3\right)$ ? What is its function equation?

Für diese Aufgabe benötigst Du folgendes Grundwissen: Lineare Funktionen
Gradient
Determine the gradient $\mathrm m$ .
$\mathrm m=\dfrac{\bigtriangleup\mathrm y}{\bigtriangleup\mathrm x}=\dfrac{{\mathrm y}_2-{\mathrm y}_1}{{\mathrm x}_2-{\mathrm x}_1}$
Insert the two points into the formula for the gradient.
$\mathrm m=\dfrac{3-(-3)}{0-2}=\dfrac 6 {-2}=-3$
Function equation
Determine the function equation.
$\mathrm y=\mathrm m\cdot\mathrm x+\mathrm t$
Insert m into the general line equation.
$\mathrm y=-3\mathrm x+\mathrm t$
Insert one of the two points into the function equation.
$3=-3\cdot0+\mathrm t$
$\mathrm t=3$
Substitute t into the function equation.
$\;\;\Rightarrow\;\;\mathrm f\left(\mathrm x\right)=-3\mathrm x+3$
7
Set up the equation of the straight line through the points $\mathrm P\left(1| 3\right)$ and $\mathrm Q\left(3|-1\right)$ .

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear Function
You are given the points $P(1\mid 3)$ and $Q(3\mid -1)$ .
You are looking for the equation of the straight line that passes through the two points.
To determine the linear equation, it is best to first consider the general form of the linear equation:
Determination of the gradient $m$
Firstly, recall the equation for the gradient of a straight line:
$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$
Plug the values $x_{1}, x_{2}, y_{1}, y_{2}$ from the points $P$ and $Q$ into the formula.
$m=\frac{-1-3}{3-1}=\ \frac{-4}{2}$
Shorten the denominator.
$m = - 2$
Now you know that the equation of the line through the points $P$ and $Q$ looks like this:
$\displaystyle y=-2\cdot x+t.$
Next, determine the y -intercept ( $t$ ).
Determination of the y -intersect $t$
To determine $t$ , insert the $x$ - and $y$ -values of one of the two points into the equation of the line. This is calculated using the point $P$ as an example
$3=-2\cdot 1+t$
$3 = - 2 + t$
$t = 5$
The $y$ -intercept of the function is 5. This gives you the function equation.
$y=-2\cdot x +5$

Set the equation of the straight line with gradient $m=-\frac43$ through the point $P (- 2 ∣ - 0.5)$ and draw it in a coordinate system.

9
Write the equation of the straight line that runs through the point $P(-3\vert3)$ and has the gradient $m = - 2$ . Draw the straight line.

Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equation
1. insert $m$ and the coordinates of the point $P$ into the general line equation and solve for $t$ .
$\displaystyle \def\arraystretch{1.25} \begin{array}{l}\begin{array}{ccccc}\mathrm y&=&\mathrm m\cdot\mathrm x+\mathrm t&&\\3&=&(-2)\cdot(-3)+\mathrm t&\;\;\;\;\;\left|-6\right.&\;\\3-6&=&\mathrm t\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;&\;\;&\;\end{array}\\\begin{array}{ccc}\;\;\;\;\;\;\;\mathrm t&=&-3\;\;\;\;\;\;\;\end{array}\\\end{array}$
2. Plug $m$ and $t$ into the general linear equation $\Rightarrow \;\;y=-2x-3$
Drawing the line
Draw the given point $P (- 3 ∣ 3)$ in the coordinate system. Then draw a slope triangle by moving 1 unit to the right and 2 units down ( $m = - 2$ ). Mark the point $Q$ there. Connect $P$ and $Q$ to form a line.
10
Set up the equation of the line through the two points and draw it.
1. $P (2 ∣ 0)$ and $Q (- 2 ∣ 2)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equation
  Determining the line equation
  $P (2 ∣ 0); Q (- 2 ∣ 2)$
  Determine the gradient $m$ of the general line equation $y=m\cdot x+t$ using the difference quotient.
  $m=\frac{2-0}{-2-2}=\frac2{-4}=-\frac12$
  Plug $m$ and the coordinates of a point, e.g. $P (2 ∣ 0)$ into the general line equation and solve for $t$ .
  $0=-\frac{1}{2}\cdot2+t\ \;\;\;\;\;\left|{+\frac12\cdot2}\right.$
  $t = 1$
  Substitute $m$ and $t$ into the general line equation.
  $\;\;\Rightarrow\;\;$ $y=-\frac12x+1$
  Drawing the line
  Draw the two points $P (2 ∣ 0)$ und $Q (- 2 ∣ 2)$ into the coordinate system and connect them to form a line.
  Do you have a question?
  Write it here 👉
2. $P (0.5 ∣ 1.5)$ and $Q (5 ∣ 3)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equation
  Determining the line equation
  $P (0.5 ∣ 1.5); Q (5 ∣ 3)$
  Determine the gradient $m$ of the general line equation $y=m\cdot x+t$ using the difference quotient.
  $m=\frac{3-1.5}{5-0.5}=\frac{1.5}{4.5}=\frac13$
  Plug $m$ and the coordinates of a point, e.g. $Q (5 ∣ 3)$ into the general line equation and solve for $t$ .
  $3=\frac13\cdot5+t \;\;\;\;\;\;\;\;|{-\frac13\cdot5}$
  $t=\frac43$
  Substitute $m$ and $t$ into the general line equation.
  $\;\;\Rightarrow\;\;$ $y=\frac13x+\frac43$
  Drawing the line
  Draw the two points $P (0.5 ∣ 1.5)$ und $Q (5 ∣ 3)$ into the coordinate system and connect them to form a line.
  Do you have a question?
  Write it here 👉
3. $P (- 2 ∣ 1)$ and $Q (6 ∣ 4)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equation
  Determining the line equation
  $P (- 2 ∣ 1)$ ; $Q (6 ∣ 4)$
  Determine the gradient $m$ of the general line equation $y=m\cdot x+t$ using the difference quotient.
  $m=\frac{4-1}{6-(-2)}=\frac38$
  Plug $m$ and the coordinates of a point, e.g. $P (- 2 ∣ 1)$ into the general line equation and solve for $t$ .
  $1=\frac38\cdot(-2)+t$
  $1= -\frac68 +t$
  $1=-\frac34+t \;\;\;\;\;\;\;\;\;|+\frac3 4$
  $t=\frac74$
  Substitute $m$ and $t$ into the general line equation.
  $\;\;\Rightarrow\;\;$ $y=\frac38x+\frac74$
  Drawing the line
  Draw the two points $P (- 2 ∣ 1)$ und $Q (6 ∣ 4)$ into the coordinate system and connect them to form a line.
  Do you have a question?
  Write it here 👉
4. $P (- 4 ∣ 1)$ and $Q (1 ∣ - 1)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equation
  Determining the line equation
  $P (- 4 ∣ 1)$ ; $Q (1 ∣ - 1)$
  Determine the gradient $m$ of the general line equation $y=m\cdot x+t$ using the difference quotient.
  $m=\frac{-1-1}{1-(-4)}=-\frac25$
  Plug $m$ and the coordinates of a point, e.g. $P (- 4 ∣ 1)$ into the general line equation and solve for $t$ .
  $1=-\frac25\cdot(-4)+t$
  $1=\frac85+t \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;|- \frac 8 5$
  $t=-\frac35$
  Substitute $m$ and $t$ into the general line equation.
  $\;\;\Rightarrow\;\;$ $y=-\frac25x-\frac35$
  Drawing the line
  Draw the two points $P (- 4 ∣ 1)$ and $Q (1 ∣ - 1)$ into the coordinate system and connect them to form a line.
  Do you have a question?
  Write it here 👉
11
Given are the y-intercept $t = 2$ and the point $P (3 ∣ - 1)$ . Calculate the corresponding equation of the line and plot the line.

Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equation
1. Insert $t$ and the coordinates of the point $P$ into the general equation of the line and solve for $m$ .
$\displaystyle \def\arraystretch{2} \begin{array}{l}\begin{array}{ccccc}\ y&=&\mathrm m\cdot\mathrm x+\mathrm t&&\\-1&=&\mathrm m\cdot3+2&\left|-2\right.&\;\\-1-2&=&\;\;\;\;\;\;\mathrm m\cdot 3\;\;\;\;\;\;\;\left|:3\right.&\;\end{array}\\\begin{array}{ccc}\;\;\;\;\;\;\;\mathrm m\;\;\;\; =\dfrac{-3}{3}=-1\;\;\;\;\;\;\;\end{array}\\\end{array}$
2. Substitute $m$ and $t$ into the general line equation $\Rightarrow\;\;y=-x+2$
Drawing the line
Draw the given point $P (3 ∣ - 1)$ in the coordinate system. Then, draw the point $S_{y} (0 ∣ 2)$ (y-intercept $t = 2)$ and connect the points $S_{y}$ and $P$ to form a line.
12
For a linear function $\mathrm h\left(\mathrm x\right)$ , suppose we know:
$\mathrm h\left(0\right)=3$ and $\mathrm h\left(-2\right)=4$ . Determine $\mathrm h\left(\mathrm x\right)$ .

Für diese Aufgabe benötigst Du folgendes Grundwissen: Line Equation
Determine the y-intercept t
For a function, the y-intercept is equal to the value at $x = 0$ .
$t = h (0) = 3$
Now use the two given points to determine the gradient $m$ using the difference quotient.
$m=\frac{3-4}{0-(-2)}=\frac{-1}{2}=-\frac12$
Put together the line equation.
$\Rightarrow y=-\frac12 x+3$

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$\displaystyle -3$	$=$	$\displaystyle m\cdot2+3$	$\displaystyle -3$
$\displaystyle -6$	$=$	$\displaystyle m\cdot2$	$\displaystyle :2$
$\displaystyle -3$	$=$	$\displaystyle m$

$\displaystyle m$	$=$	$\displaystyle \frac{8-7}{-3-5}$
	↓	Simplify numerator and denominator
$\displaystyle m$	$=$	$\displaystyle \frac{1}{-8}$
$\displaystyle m$	$=$	$\displaystyle - \frac18$

$\displaystyle m$	$=$	$\displaystyle \frac{4-2}{3-1}$
	↓	Simplify numerator and denominator
$\displaystyle m$	$=$	$\displaystyle \frac22$
$\displaystyle m$	$=$	$\displaystyle 1$

$\displaystyle 3$	$=$	$\displaystyle -2\cdot1+t$
$\displaystyle 3$	$=$	$\displaystyle -2+t$	$\displaystyle +2$
$\displaystyle t$	$=$	$\displaystyle 5$

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

Exercises: Finding line equations

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Linear function f(x)f(x)f(x)

Linear function g(x)g(x)g(x)

Linear function h(x)h(x)h(x)

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Linear function f(x)f(x)f(x)

Linear function g(x)g(x)g(x)

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Gradient

Function equation

Determination of the gradient mmm

Determination of the y -intersect ttt

Setting up line equations

Drawing the line

Drawing the line

Determining the line equation

Drawing the line

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Determining the line equation

Drawing the line

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Determining the line equation

Drawing the line

Do you have a question?

Determining the line equation

Drawing the line

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Drawing the line

Linear function $f (x)$

Linear function $g (x)$

Linear function $h (x)$

Linear function $f (x)$

Linear function $g (x)$

Determination of the gradient $m$

Determination of the y -intersect $t$