Exercises: Intersection of two lines

Bestimme den Schnittpunkt beider Geraden und zeichne diesen in ein Koordinatensystem.

$f\left(x\right)=-3x+\frac54;\;g\left(x\right)=-x-1$

$f:\;2y-x=3;\;g\left(x\right)=-\frac12x+4$

$f\left(x\right)=-\frac23x-1;\;g\left(x\right)=\frac16x-4$

$f:\;x=2;\;g\left(x\right)=-\frac34x-\frac32$

Determine the intersection of the two lines and draw the graphs in a coordinate system.

$f\left(x\right)=0.05x+20;\;\;\;\;\;g\left(x\right)=0.15x+15$

Calculate the intersection points of a straight line.

Consider are the function equations of the two lines $g_{1} (x)$ and $g_2\left(x\right)$ . Calculate the intersection of the two lines and draw the lines in a coordinate system.

${g}_1\left(x\right)=\frac12x+2\;\;\;\;\;{g}_2\left(x\right)=-\frac12x+4$

${g}_1\left(x\right)=2x-1\;\;\;\;\;{g}_2\left(x\right)=-2x+1$

${g}_1\left(x\right)=\frac34x-4\;\;\;\;\;{g}_2\left(x\right)=-\frac12x-1$

${g}_1\left(x\right)=-\frac12x+2\;\;\;\;\;{g}_2\left(x\right)=\frac12x+3$

${g}_1\left(x\right)=\frac23x+2\;\;\;\;\;{g}_2\left(x\right)=\frac12x+3$

${g}_1\left(x\right)=\frac34x+1\;\;\;\;\;{g}_2\left(x\right)=\frac12x+2$

Consider the following graphs.

Determine the function equations of all 4 lines.

Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
Line equation of $f$
In order to determine the line equation of $f$ , first read off two points from the diagram that lie on the straight line $f$ . In the above example, you may for instance get $A (0 ∣ 3)$ and $B (4 ∣ 2)$ . Use these two points to determine the slope of $f$ via the difference quotient
$\displaystyle m_f=\frac{y_B-y_A}{x_B-x_A}$
Plug in the values
$\displaystyle m_f=\frac{2-3}{4-0}=-\frac14$
Now determine the $y$ -axis intercept $b_{f}$ by plugging in any point on $f$ into the general line equation. Alternatively, you may directly read off at which value $f$ intersects the $y$ -axis (if possible).
$f (x) : y = m_{f} x + b_{f}$
For instance, plug in $A$ .
$\displaystyle 3=-\frac14\cdot0+b_f$
Simplify.
$3=b_f\Rightarrow b_f=3$
So the line equation is $\displaystyle f(x)=-\frac14\cdot x+3$ .
Line equation of $g$
In order to determine the line equation of $g$ , first read off two points from the diagram that lie on the straight line $g$ . In the above example, you may for instance get $C (- 4 ∣ 0)$ and $D (0 ∣ 1)$ . Use these two points to determine the slope of $g$ via the difference quotient
$\displaystyle m_g=\frac{y_D-y_C}{x_D-x_C}$
Plug in the values
$\displaystyle m_g=\frac{1-0}{0-(-4)}=\frac14$
Now determine the $y$ -axis intercept $b_{g}$ by plugging in any point on $g$ into the general line equation. Alternatively, you may directly read off at which value $g$ intersects the $y$ -axis (if possible).
$g (x) : y = m_{g} x + b_{g}$
For instance, plug in $D$ .
$\displaystyle 1=\frac14\cdot0+b_g$
Simplify.
$\displaystyle 1=b_g\Rightarrow b_g=1$
So the line equation is $\displaystyle g(x)=\frac14\cdot x+1$ .
Line equation of $h$
In order to determine the line equation of $h$ , first read off two points from the diagram that lie on the straight line $h$ . In the above example, you may for instance get $E (- 1 ∣ 0)$ and $A (0 ∣ 3)$ . Use these two points to determine the slope of $h$ via the difference quotient
$\displaystyle m_h=\frac{y_A-y_E}{x_A-x_E}$
Plug in the values
$\displaystyle m_h=\frac{3-0}{0-(-1)}=3$
Now determine the $y$ -axis intercept $b_{h}$ by plugging in any point on $h$ into the general line equation. Alternatively, you may directly read off at which value $h$ intersects the $y$ -axis (if possible).
$h (x) : y = m_{h} x + b_{h}$
For instance, plug in $A$ .
$3=3\cdot0+b_h$
Simplify.
$3=b_h\Rightarrow b_h=3$
So the line equation is $\displaystyle h(x)=3\cdot x+3$ .
Line equation of $i$
In order to determine the line equation of $g$ , first read off two points from the diagram that lie on the straight line $g$ . In the above example, you may for instance get $F (0 ∣ - 3)$ and $S (6 ∣ 0)$ . Use these two points to determine the slope of $g$ via the difference quotient
$\displaystyle m_i=\frac{y_S-y_F}{x_S-x_F}$
Plug in the values
$\displaystyle m_i=\frac{0-(-3)}{6-0}=\frac12$
Now determine the $y$ -axis intercept $b_{i}$ by plugging in any point on $i$ into the general line equation. Alternatively, you may directly read off at which value $i$ intersects the $y$ -axis (if possible).
$i (x) : y = m_{i} x + b_{i}$
For instance, plug in $F$ .
$\displaystyle-3=\frac12\cdot0+b_i$
Simplify.
$-3=b_i\Rightarrow b_i=-3$
So the line equation is $\displaystyle i(x)=\frac12\cdot x-3$ .
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Determine the intersection of g and h , and the zero of f.

Calculate the two intersections that lie outside the figure.

The intersection of $h$ and $i$ and the intersection of $g$ and $i$ lie outside the figure above.

Intersection point $T (x_{T} ∣ y_{T})$ of $h$ and $i$

To determine the point of intersection of two functions, you set them equal and solve for $x$ . The function equations (determined in the previous problem) are $h (x) : y = 3 x + 3$ and $\displaystyle i(x):y=\frac12x-3$ .

$\displaystyle \displaystyle3x_T+3$	$=$	$\displaystyle \frac{1}{2}x_T-3$	$\displaystyle -\frac{1}{2}x-3$
$\displaystyle \frac{5}{2}x_T$	$=$	$\displaystyle -6$	$\displaystyle :\frac{5}{2}$
$\displaystyle x_T$	$=$	$\displaystyle -\frac{12}{5}$

Now substitute $-\frac{12}5$ into the line equation of $h$ or $i$ to determine $y_{T}$ .

$h (x_{T}) : y_{T} = 3 \cdot x_{T} + 3 h(x_T):y_T=3⋅x_T+3$

Plug in $x_{T}$ .

$\displaystyle y_T=3\cdot\left(-\frac{12}5\right)+3=-\frac{21}5$

The lines $h$ and $i$ therefore intersect at $\displaystyle T\left(-\frac52\left|-\frac{21}5\right.\right)$ .

Intersection point $Q (x_{Q} ∣ y_{Q})$ of $g$ and $i$

To determine the point of intersection of two functions, you set them equal and solve for $x$ . The function equations (determined in the previous problem) are $\displaystyle g(x):y=\frac14x+1$ and $\displaystyle i(x):y=\frac12x-3$ .

$\displaystyle \displaystyle\frac14x_Q+1$	$=$	$\displaystyle \frac{1}{2}x_Q-3$	$\displaystyle -\frac{1}{2}x-1$
$\displaystyle \displaystyle -\frac14x_Q$	$=$	$\displaystyle -4$	$\displaystyle :\left(-\frac{1}{4}\right)$
$\displaystyle x_Q$	$=$	$\displaystyle 16$

Now substitute $16$ into the line equation of $g$ or $i$ to determine $y_{Q}$ .

$\displaystyle g(x_Q):y_Q=\frac14⋅x_Q+1$

Plug in $x_{Q}$ .

$\displaystyle y_Q=\frac14\cdot16+1=5$

The lines $g$ and $i$ therefore intersect at $\displaystyle Q(16|5)$ .

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What is the maximum number of intersections of four straight lines?
If no pair of lines is parallel, then there are a total of 6 intersections, namely the following:
$f$ and $g$
$f$ and $h$
$f$ and $i$
$g$ and $h$
$g$ and $i$
$h$ and $i$
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Calculate the intersection of the pairs of lines.

$y = 3 x + 4$ and $y = - 2 x + 14$

$y = 6 x - 3$ and $y = 7 x - 11$

$y = 8 x + 3$ and $y = - 4 x + 6$

$y = 7 x - 14$ and $y = 7 x - 3$

Geraden mit gleicher Steigung (hier 7) schneiden sich nicht, denn sie sind parallel zueinander.Setzt man die Funktionsterme gleich, so erhält man eine falsche Aussage, also keinen Schnittpunkt.
$\displaystyle 7x-14$ $=$ $\displaystyle 7x-3$ $\displaystyle -7x$
$\displaystyle -14$ $=$ $\displaystyle -3$
↓
False statement!
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$y=\frac{1}{6}x-4$ and $y=\frac{1}{3}x-10$

$y=\frac{1}{2}x+\frac{3}{2}$ and $y=\frac{1}{2}$

6
Show by calculation that the three straight lines $g_{1} : y = 0.5 x$ ; $g_{2} : y = x - 1.5$ and $g_{3} : y = - 2 x + 7.5$ intersect at exactly one point.

$g_{2}$ : $y = x - 1.5$ ; $g_{3}$ : $y = - 2 x + 7.5$
Es ist hier zu empfehlen, zunächst den Schnittpunkt von zwei Geraden zu berechnen und dann zu prüfen, ob der Schnittpunkt auch ein Punkt auf der anderen Gerade ist.
Zwei der drei Geraden werden gleichgesetzt um die x-Koordinate des Schnittpunktes zu berechnen.
$\displaystyle x-1.5$ $=$ $\displaystyle -2x+7.5$
↓
First add 2x and then 1.5.
$\displaystyle 3x$ $=$ $\displaystyle 9$
↓
divide by 3.
$\displaystyle x$ $=$ $\displaystyle 3$
$=$
Insert the $x$ -value into one of the two line equations (e.g. that of $g_{2}$ ) to calculate the $y$ -coordinate of the intersection of $g_{2}$ and $g_{3}$ .
$y = 3 - 1.5 = 1.5$
$\Rightarrow$ $S (3 ∣ 1.5)$
Now check whetehr the intersection point is also in the third line:
$g_{1}$ : $y = 0.5 x$
The point of intersection is now inserted into $g_{1}$ . This means that $x$ and $y$ of the $g_{1}$ are replaced by 3 and 1.5.
$1.5=0.5\cdot3$
Check whether the resulting statement is true in order to determine whether $g_{1}$ runs through $S$ .
$1.5 = 1.5$
$\Rightarrow$ This is a true statement, so $S$ is the common and only intersection of all three straight lines.

Check whether the following lines $g, h$ and $i$ run through a common point.

$g\left(x\right)=x+1;\;\;\;\;\;h:\;2y+x+4=0;\;\;\;\;\;3y-5x=7$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function

First bring all lines equations into the general form $y = m x + t$ .

Line $g$ :

$g (x) = x + 1$

$\Leftrightarrow y=x+1$

Line $h$ :

$2 y + x + 4 = 0$

$\Leftrightarrow 2y=-x-4$

$\Leftrightarrow y=-\frac12x-2$

Line $i$ :

$3 y - 5 x = 7$

$\Leftrightarrow 3y=5 x+7$

$\Leftrightarrow y=\frac{5}{3}x+\frac{7}{3}$

Determine the intersection of $g$ and $h$

Set the respective right sides equal.

$\displaystyle x+1$	$=$	$\displaystyle -\frac{1}{2}x-2$	$\displaystyle +\frac{1}{2}x-1$
	↓	Get all $x$ -terms on one side.
$\displaystyle x+\frac{1}{2}x$	$=$	$\displaystyle -2-1$
	↓	Conclude
$\displaystyle \frac{3}{2}x$	$=$	$\displaystyle -3$	$\displaystyle :\frac{3}{2}$
$\displaystyle x$	$=$	$\displaystyle -2$
	↓	Plug into $g$ to get the $y$ -coordinate.
$\displaystyle y$	$=$	$\displaystyle -2+1\ =\ -1$

The intersection point is $S_{gh}(-2\,|\,-1).$

Determine the intersection of $g$ and $i$

Set the respective right sides equal.

$\displaystyle x+1$	$=$	$\displaystyle \frac{5}{3}x+\frac{7}{3}$	$\displaystyle -\frac{5}{3}x-1$
	↓	Get all $x$ -terms on one side.
$\displaystyle x-\frac{5}{3}x$	$=$	$\displaystyle \frac{7}{3}-1$
	↓	Conclude.
$\displaystyle -\frac{2}{3}x$	$=$	$\displaystyle \frac{4}{3}$	$\displaystyle :(-\frac{2}{3})$
$\displaystyle x$	$=$	$\displaystyle -2$
	↓	Plug into $g$ to get the $y$ -coordinate.
$\displaystyle y$	$=$	$\displaystyle -2+1\ =\ -1$

The intersection point is $S_{gi}(-2\,|\,-1).$

Since $g$ intersects with $h$ and with $i$ at the same point, $h$ and $i$ also intersect at this point.

The lines therefore all run through the common point $(- 2 ∣ - 1)$ .

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$g\left(x\right)=\frac16 x+\frac32;\;\;\;\;\; h\left( x\right)=-\frac23 x+2;\;\;\;\;\; i:\;2 x- y=3$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function

First bring all lines equations into the general form $y = m x + t$ .

Line $g$ :

$g(x)=\frac{1}{6}x+\frac{3}{2}$

$\Leftrightarrow y=\frac16x+\frac32$

Line $h$ :

$h(x)=-\frac23x+2$

$\Leftrightarrow y=-\frac23x+2$

Line $i$ :

$2 x - y = 3$

$\Leftrightarrow y=2 x-3$

Determine the intersection of $g$ and $h$

Set the respective right sides equal.

$\displaystyle \frac{1}{6}x+\frac{3}{2}$	$=$	$\displaystyle -\frac{2}{3}x+2\quad$	$\displaystyle +\frac{2}{3}x-\frac{3}{2}$
	↓	Get all $x$ -terms on one side.
	$=$
	↓	Conclude.
$\displaystyle \frac{1}{6}x+\frac{2}{3}x$	$=$	$\displaystyle 2-\frac{3}{2}$
$\displaystyle \frac{5}{6}x$	$=$	$\displaystyle \frac{1}{2}$	$\displaystyle :\frac{5}{6}$
$\displaystyle x$	$=$	$\displaystyle \frac{3}{5}$
	↓	Plug into $g$ to get the $y$ -coordinate.
$\displaystyle y$	$=$	$\displaystyle \frac{1}{6}\cdot\frac{3}{5}+\frac{3}{2}$
$\displaystyle y$	$=$	$\displaystyle \frac{1}{10}+\frac{3}{2}\ =\ \frac{16}{10}$
$\displaystyle y$	$=$	$\displaystyle \frac{8}{5}$

The intersection point is $S_{gh}\left(\frac35\,|\,\frac85\right).$

Determine the intersection of $g$ and $i$

Set the respective right sides equal.

$\displaystyle \frac{1}{6}x+\frac{3}{2}$	$=$	$\displaystyle 2x-3$	$\displaystyle -2x-\frac{3}{2}$
	↓	Get all $x$ -terms on one side.
$\displaystyle \frac{1}{6}x-2x$	$=$	$\displaystyle -3-\frac{3}{2}$
	↓	Conclude.
$\displaystyle -\frac{11}{6}x$	$=$	$\displaystyle -\frac{9}{2}$	$\displaystyle :\left(-\frac{11}{6}\right)$
$\displaystyle x$	$=$	$\displaystyle \frac{9}{2}\cdot\frac{6}{11}=\frac{27}{11}$
	↓	Plug into $g$ to get the $y$ -coordinate.
$\displaystyle y$	$=$	$\displaystyle \frac{1}{6}\cdot\frac{27}{11}+\frac{3}{2}$
$\displaystyle y$	$=$	$\displaystyle \frac{9}{22}+\frac{3}{2\ }\ =\ \frac{42}{44}$
$\displaystyle y$	$=$	$\displaystyle \frac{21}{11}$

The intersection point is $S_{gi}\left(\frac{27}{11}\,|\,\frac{21}{11}\right).$

Thus the line $g$ intersects the line $h$ at a different point than the line $i$ . So the lines do not run through a common point.

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8
Determination of intersection points
Consider the following two lines $g$ and $h$ .
1. Determine the line equations of $g$ and $h$ .
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
  $\displaystyle y=mx+t$
  The parameter $m$ is the slope of the line. The parameter $t$ is the $y$ -axis intercept.
  Line $g$ :
  $y = m_{g} x + t_{g}$
  The line $g$ has slope $-\frac{3}{2}$ , so $m_g = -\frac{3}{2}$ .
  $y=-\frac{3}{2}x+t_g$
  It intersects the $y$ -axis at $\frac{9}{2}$ , so $t_g = \frac{9}{2}$ .
  $y=-\frac{3}{2}x+\frac{9}{2}$
  Line $h$ :
  $y = m_{h} x + t_{h}$
  The line $h$ has slope $2$ , so $m_g = -\frac{3}{2}$ $m_{h} = 2$
  $y = 2 x + t_{h}$
  It intersects the $y$ -axis at $1$ , so $t_{h} = 1$ .
  $y = 2 x + 1$
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2. Read off the intersection point.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Lines in coordinate systems
  Reading off the intersection point
  From the plot, we may read off the intersection point $\text{SP}(1|3)$ .
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$\displaystyle -3{x}+\frac{5}{4}$	$=$	$\displaystyle -x-1$	$\displaystyle +3{x}+1$
$\displaystyle -{x}+3{x}$	$=$	$\displaystyle \frac{5}{4}+1$
	↓	add
$\displaystyle 2x$	$=$	$\displaystyle \frac{9}{4}$	$\displaystyle :2$
$\displaystyle x$	$=$	$\displaystyle \frac{9}{8}$

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

$\displaystyle \frac{3}{2}+\frac{{x}}{2}$	$=$	$\displaystyle -\frac{1}{2}x+4$	$\displaystyle +\frac{{x}}{2}-1.5$
$\displaystyle \frac{{x}}{2}+\frac{{x}}{2}$	$=$	$\displaystyle 4-1.5$
	↓	add and subtract
$\displaystyle x$	$=$	$\displaystyle 2.5$

$\displaystyle 0.05{x}+20$	$=$	$\displaystyle 0.15{x}+15$	$\displaystyle -0.05{x}-15$
$\displaystyle 20-15$	$=$	$\displaystyle 0.15{x}-0.05{x}$
	↓	Subtrahiere und vertausche die Seiten.
$\displaystyle 0.10{x}$	$=$	$\displaystyle 5$	$\displaystyle :0.1$
$\displaystyle x$	$=$	$\displaystyle \frac{5}{0.1}$
	↓	Dividiere.
$\displaystyle {x}_S$	$=$	$\displaystyle 50$

$\displaystyle \frac{1}{2}{x}+2$	$=$	$\displaystyle -\frac12x+4\\$	$\displaystyle \left\|+\frac12x-2\right.$
$\displaystyle \frac{1}{2}{x}+\frac{1}{2}{x}$	$=$	$\displaystyle 4-2$
$\displaystyle {x}_S$	$=$	$\displaystyle 2$
	↓	Plug $x = x_{s}$ into ${g}_1\left( x\right)$ .
$\displaystyle {y}$	$=$	$\displaystyle 1+2$
$\displaystyle {y}_S$	$=$	$\displaystyle 3$

$\displaystyle y$	$=$	$\displaystyle -\frac{9}{8}-1$
	↓	subtract
$\displaystyle y$	$=$	$\displaystyle -2.125$

$\displaystyle y$	$=$	$\displaystyle \frac{3}{2}+\frac{2.5}{2}$
	↓	add
$\displaystyle y$	$=$	$\displaystyle 2.75$

$\displaystyle -\frac{2}{3}{x}-1$	$=$	$\displaystyle \frac{1}{6}{x}-4$	$\displaystyle +\frac{2}{3}{x}+4$
$\displaystyle \frac{1}{6}{x}+\frac{2}{3}{x}$	$=$	$\displaystyle -1+4$
	↓	Addiere.
$\displaystyle \frac{5}{6}{x}$	$=$	$\displaystyle 3$	$\displaystyle :\frac{5}{6}$
$\displaystyle x$	$=$	$\displaystyle 3\;:\;\frac56$
	↓	divide by a fraction $\rightarrow$ multiply by the inverse
$\displaystyle x$	$=$	$\displaystyle 3\; \cdot\;\frac65$
	↓	multiply
$\displaystyle x$	$=$	$\displaystyle \frac{18}{5}$

$\displaystyle y$	$=$	$\displaystyle \frac{1}{6}\cdot\frac{18}{5}-4$
	↓	multiply
$\displaystyle y$	$=$	$\displaystyle \frac{3}{5}-4$
	↓	subtract
$\displaystyle y$	$=$	$\displaystyle -3.4$

$\displaystyle y$	$=$	$\displaystyle -\frac{3}{4}\cdot2-\frac{3}{2}$
	↓	multiply
$\displaystyle y$	$=$	$\displaystyle -\frac{3}{2}-\frac{3}{2}$
	↓	subtract
$\displaystyle y$	$=$	$\displaystyle -\frac{6}{2}$
	↓	divide
$\displaystyle y$	$=$	$\displaystyle -3$

$\displaystyle y$	$=$	$\displaystyle 0.05\cdot50+20$
	↓	Multipliziere.
$\displaystyle y$	$=$	$\displaystyle 2.5+20$
	↓	Addiere.
$\displaystyle {y}_S$	$=$	$\displaystyle 22.5$

$\displaystyle 2x-1$	$=$	$\displaystyle -2x+1$	$\displaystyle +2x+1$
$\displaystyle 2x+2x$	$=$	$\displaystyle 1+1$
$\displaystyle 4x$	$=$	$\displaystyle 2$	$\displaystyle :4$
$\displaystyle \mathrm{x}_S$	$=$	$\displaystyle \frac{1}{2}$
	↓	Plug $x = x_{s}$ into ${g}_1\left( x\right)$ .
$\displaystyle y$	$=$	$\displaystyle 2\cdot\frac{1}{2}-1$
$\displaystyle y$	$=$	$\displaystyle 1-1$
$\displaystyle \mathrm{y}_S$	$=$	$\displaystyle 0$

$\displaystyle g_1(x)$	$=$	$\displaystyle g_2(x)$
$\displaystyle \frac{3}{4}x-4$	$=$	$\displaystyle -\frac{1}{2}x-1$	$\displaystyle +\frac{1}{2}x+4$
$\displaystyle \frac{3}{4}x+\frac{1}{2}x$	$=$	$\displaystyle -1+4$
$\displaystyle 1.25x$	$=$	$\displaystyle 3$	$\displaystyle 1.25$
$\displaystyle x$	$=$	$\displaystyle \frac{3}{1.25}$
$\displaystyle {x}_S$	$=$	$\displaystyle 2.4$
	↓	Plug $x = x_{s}$ into ${g}_1\left( x\right)$ .
$\displaystyle y$	$=$	$\displaystyle \frac{3}{4}\cdot2.4-4$
$\displaystyle y$	$=$	$\displaystyle 1.8-4$
$\displaystyle {y}_S$	$=$	$\displaystyle -2.2$

$\displaystyle g_1\left(x\right)$	$=$	$\displaystyle g_2\left(x\right)$
$\displaystyle -\frac{1}{2}{x}+2$	$=$	$\displaystyle \frac12x+3\\$	$\displaystyle \left\|+\frac12x\right.-3$
$\displaystyle 2-3$	$=$	$\displaystyle \frac{1}{2}x+\frac{1}{2}x$
$\displaystyle {x}_S$	$=$	$\displaystyle -1$
	↓	Plug $x = x_{s}$ into ${g}_1\left( x\right)$ .
$\displaystyle y$	$=$	$\displaystyle -\frac12\cdot\left(-1\right)+2\\$
$\displaystyle y$	$=$	$\displaystyle \frac{1}{2}+2$
$\displaystyle {y}_S$	$=$	$\displaystyle 2.5$

$\displaystyle g_1(x)$	$=$	$\displaystyle g_2(x)$
$\displaystyle \frac23x+2$	$=$	$\displaystyle \frac12x+3$	$\displaystyle -\frac{1}{2}x-2$
	↓	Get the fractions on a common denominator.
$\displaystyle \frac{2}{3}x-\frac{1}{2}x$	$=$	$\displaystyle 3-2$
	↓	subtract
$\displaystyle \frac{4}{6}x-\frac{3}{6}x$	$=$	$\displaystyle 3-2$
$\displaystyle \frac{1}{6}x$	$=$	$\displaystyle 1$	$\displaystyle :\frac{1}{6}$
	↓	dividing by a fraction $\rightarrow$ multiplying with the inverse
$\displaystyle x$	$=$	$\displaystyle \frac{1}{\frac{1}{6}}$
$\displaystyle x_S$	$=$	$\displaystyle 6$
	↓	Plug $x = x_{s}$ into ${g}_1\left( x\right)$ .
$\displaystyle y$	$=$	$\displaystyle \frac{2}{3}\cdot6+2$
	↓	add
$\displaystyle y$	$=$	$\displaystyle 4+2$
$\displaystyle y_S$	$=$	$\displaystyle 6$

$\displaystyle g_1(x)$	$=$	$\displaystyle g_2(x)$
$\displaystyle \frac{3}{4}x+1$	$=$	$\displaystyle \frac{1}{2}x+2$	$\displaystyle -\frac{1}{2}x-1$
$\displaystyle \frac{3}{4}x-\frac{1}{2}x$	$=$	$\displaystyle 2-1$
	↓	Get the fractions on a common denominator.
$\displaystyle \frac{3}{4}x-\frac{2}{4}x$	$=$	$\displaystyle 2-1$
	↓	subtract
$\displaystyle \frac{1}{4}x$	$=$	$\displaystyle 1$	$\displaystyle :\frac{1}{4}$
	↓	dividing by a fraction $\rightarrow$ multiplying with the inverse
$\displaystyle x$	$=$	$\displaystyle \frac{1}{\frac{1}{4}}$
$\displaystyle x_S$	$=$	$\displaystyle 4$
	↓	Plug $x = x_{s}$ into ${g}_1\left( x\right)$ .
$\displaystyle y$	$=$	$\displaystyle \frac{3}{4}\cdot4+1$
	↓	shorten
$\displaystyle y$	$=$	$\displaystyle 3+1$
	↓	add
$\displaystyle y_S$	$=$	$\displaystyle 4$

$\displaystyle \displaystyle\frac14x_P+1$	$=$	$\displaystyle 3x_P+3$	$\displaystyle -3x_p-1$
	↓	subtract $3 x_{P}$ and $1$ .
$\displaystyle \displaystyle -\frac{11}{4}x_P$	$=$	$\displaystyle 2$	$\displaystyle \div\left(-\frac{11}{4}\right)$
	↓	divide by $-\frac{11}4$ .
$\displaystyle x_p$	$=$	$\displaystyle -\frac{8}{11}$
	$=$

$\displaystyle \displaystyle-\frac14x_{N_f}+3$	$=$	$\displaystyle 0$	$\displaystyle -3$
$\displaystyle \displaystyle -\frac14x_{N_f}$	$=$	$\displaystyle -3$	$\displaystyle :\frac{1}{4}$
$\displaystyle \displaystyle x_{N_f}$	$=$	$\displaystyle 12$

$\displaystyle y$	$=$	$\displaystyle 3x+4$
$\displaystyle y$	$=$	$\displaystyle -2x+14$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\displaystyle 3x+4$	$=$	$\displaystyle -2x+14$	$\displaystyle +2x;\ -4$
$\displaystyle 5x$	$=$	$\displaystyle 10$	$\displaystyle :5$
$\displaystyle x$	$=$	$\displaystyle 2$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$\displaystyle y$	$=$	$\displaystyle 3\cdot2+4$
	↓	First multiply, then divide.
$\displaystyle y$	$=$	$\displaystyle 10$
	↓	$\;\;\Rightarrow\;\;$ Intersection point: $S (2 ∣ 10)$

$\displaystyle y$	$=$	$\displaystyle 6x-3$
$\displaystyle y$	$=$	$\displaystyle 7x-11$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\displaystyle 6x-3$	$=$	$\displaystyle 7x-11$	$\displaystyle +3;\ -7x\$
$\displaystyle -x$	$=$	$\displaystyle -8$	$\displaystyle :\left(-1\right)$
$\displaystyle x$	$=$	$\displaystyle 8$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$\displaystyle y$	$=$	$\displaystyle 6\cdot8-3$
	↓	First multiply, then divide.
$\displaystyle y$	$=$	$\displaystyle 45$
	↓	$\Rightarrow$ Intersection point: $S (8 ∣ 45)$

$\displaystyle y$	$=$	$\displaystyle 8x+3$
$\displaystyle y$	$=$	$\displaystyle -4x+6$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\displaystyle 8x+3$	$=$	$\displaystyle -4x+6$	$\displaystyle -3;\ +4x$
$\displaystyle 12x$	$=$	$\displaystyle 3$	$\displaystyle :12$
$\displaystyle x$	$=$	$\displaystyle \frac{3}{12}=\frac{1}{4}$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$\displaystyle y$	$=$	$\displaystyle 8\cdot\frac{1}{4}+3$
	↓	First multiply, then divide.
$\displaystyle y$	$=$	$\displaystyle 5$
	↓	$\Rightarrow$ Intersection point: $S(\frac{1}{4}\|5)$

$\displaystyle 7x-14$	$=$	$\displaystyle 7x-3$	$\displaystyle -7x$
$\displaystyle -14$	$=$	$\displaystyle -3$
	↓	False statement!

$\displaystyle y$	$=$	$\displaystyle \frac{1}{6}x-4$
$\displaystyle y$	$=$	$\displaystyle \frac{1}{3}x-10$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\displaystyle \frac{1}{6}x-4$	$=$	$\displaystyle \frac{1}{3}x-10$	$\displaystyle -\frac{1}{3}x;+4$
$\displaystyle -\frac{1}{6x}$	$=$	$\displaystyle -6$	$\displaystyle :\left(-\frac{1}{6}\right)$
$\displaystyle x$	$=$	$\displaystyle 36$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$\displaystyle y$	$=$	$\displaystyle \frac{1}{3}\cdot36-10$
	↓	First multiply, then divide.
$\displaystyle y$	$=$	$\displaystyle 2$
	↓	$\Rightarrow$ Intersection point: $S (36 ∣ 2)$

$\displaystyle y$	$=$	$\displaystyle \frac{1}{2}x+\frac{3}{2}$
$\displaystyle y$	$=$	$\displaystyle \frac{1}{2}$
	↓	The functions are set equal to calculate the $x$ -coordinate of the intersection.
$\displaystyle \frac{1}{2}x+\frac{3}{2}$	$=$	$\displaystyle \frac{1}{2}$	$\displaystyle -\frac{3}{2}$
$\displaystyle \frac{1}{2}x$	$=$	$\displaystyle -1$	$\displaystyle \cdot2$
$\displaystyle x$	$=$	$\displaystyle -2$
	↓	Insert the $x$ -value into one of the two functions to calculate the $y$ -coordinate
$\displaystyle y$	$=$	$\displaystyle \frac{1}{2}$
	↓	$\Rightarrow$ Intersection point: $S (- 2 ∣ 0.5)$

$\displaystyle x-1.5$	$=$	$\displaystyle -2x+7.5$
	↓	First add 2x and then 1.5.
$\displaystyle 3x$	$=$	$\displaystyle 9$
	↓	divide by 3.
$\displaystyle x$	$=$	$\displaystyle 3$
	$=$

Exercises: Intersection of two lines

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Determining the intersection point

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Calculating the line intersection

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Calculating the line intersection

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Calculating the line intersection

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Calculating the line intersection

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Calculating the line intersection

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Calculating the line intersection

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Line equation of fff

Line equation of ggg

Line equation of hhh

Line equation of iii

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Intersection point P(xp∣yp)P(x_p|y_p)P(xp​∣yp​) of ggg and hhh

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Intersection point T(xT∣yT)T(x_T|y_T)T(xT​∣yT​) of hhh and iii

Intersection point Q(xQ∣yQ)Q(x_Q|y_Q)Q(xQ​∣yQ​) of ggg and iii

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Determine the intersection of ggg and hhh

Determine the intersection of ggg and iii

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Determine the intersection of ggg and hhh

Determine the intersection of ggg and iii

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Determination of intersection points

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Reading off the intersection point

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Line equation of $f$

Line equation of $g$

Line equation of $h$

Line equation of $i$

Intersection point $P (x_{p} ∣ y_{p})$ of $g$ and $h$

Intersection point $T (x_{T} ∣ y_{T})$ of $h$ and $i$

Intersection point $Q (x_{Q} ∣ y_{Q})$ of $g$ and $i$

Determine the intersection of $g$ and $h$

Determine the intersection of $g$ and $i$

Determine the intersection of $g$ and $h$

Determine the intersection of $g$ and $i$