Exercises: Intersection of two lines

1
Bestimme den Schnittpunkt beider Geraden und zeichne diesen in ein Koordinatensystem.
$f (x) = - 3 x + \frac{5}{4}; g (x) = - x - 1$

For this task you need the following basic knowledge: Linear function
Set the functions $f (x)$ and $g (x)$ equal to obtain the intersection point.
$- 3 x + \frac{5}{4}$ $=$ $- x - 1$ $+ 3 x + 1$
$- x + 3 x$ $=$ $\frac{5}{4} + 1$
↓
add
$2 x$ $=$ $\frac{9}{4}$ $: 2$
$x$ $=$ $\frac{9}{8}$
Insert $x$ into one of the two lines equations, for example $g$ :
$y$ $=$ $- \frac{9}{8} - 1$
↓
subtract
$y$ $=$ $- 2.125$
$\Rightarrow S (1.25 / - 2.125)$
Sketch

$f : 2 y - x = 3; g (x) = - \frac{1}{2} x + 4$

For this task you need the following basic knowledge: Linear function
Set the functions $f (x)$ and $g (x)$ equal to obtain the intersection point.
First solve the function $f$ for $y$ .
$2 y - x$ $=$ $3$ $+ x$
$2 y$ $=$ $3 + x$ $: 2$
$y$ $=$ $\frac{3}{2} + \frac{x}{2}$
Now set f and g equal
$\frac{3}{2} + \frac{x}{2}$ $=$ $- \frac{1}{2} x + 4$ $+ \frac{x}{2} - 1.5$
$\frac{x}{2} + \frac{x}{2}$ $=$ $4 - 1.5$
↓
add and subtract
$x$ $=$ $2.5$
Plug $x$ into one of the two line equations, for example $f$ :
$y$ $=$ $\frac{3}{2} + \frac{2.5}{2}$
↓
add
$y$ $=$ $2.75$
$\Rightarrow S (2.5 / 2.75)$

$f (x) = - \frac{2}{3} x - 1; g (x) = \frac{1}{6} x - 4$

For this task you need the following basic knowledge: Linear function
Set the functions $f (x)$ and $g (x)$ equal to obtain the intersection point.
$- \frac{2}{3} x - 1$ $=$ $\frac{1}{6} x - 4$ $+ \frac{2}{3} x + 4$
$\frac{1}{6} x + \frac{2}{3} x$ $=$ $- 1 + 4$
↓
Addiere.
$\frac{5}{6} x$ $=$ $3$ $: \frac{5}{6}$
$x$ $=$ $3 : \frac{5}{6}$
↓
divide by a fraction $\to$ multiply by the inverse
$x$ $=$ $3 \cdot \frac{6}{5}$
↓
multiply
$x$ $=$ $\frac{18}{5}$
Insert $x$ into one of the two lines equations, for example $g$ :
$y$ $=$ $\frac{1}{6} \cdot \frac{18}{5} - 4$
↓
multiply
$y$ $=$ $\frac{3}{5} - 4$
↓
subtract
$y$ $=$ $- 3.4$
$\Rightarrow S (3.6 / - 3.4)$

$f : x = 2; g (x) = - \frac{3}{4} x - \frac{3}{2}$

For this task you need the following basic knowledge: Linear function
Plug $x = 2$ into $g (x)$
$y$ $=$ $- \frac{3}{4} \cdot 2 - \frac{3}{2}$
↓
multiply
$y$ $=$ $- \frac{3}{2} - \frac{3}{2}$
↓
subtract
$y$ $=$ $- \frac{6}{2}$
↓
divide
$y$ $=$ $- 3$
$\Rightarrow S (2 / - 3)$
2
Determine the intersection of the two lines and draw the graphs in a coordinate system.
$f (x) = 0.05 x + 20; g (x) = 0.15 x + 15$

For this task you need the following basic knowledge: Linear function
Determining the intersection point
Set $f (x)$ and $g (x)$ equal.
$0.05 x + 20$ $=$ $0.15 x + 15$ $- 0.05 x - 15$
$20 - 15$ $=$ $0.15 x - 0.05 x$
↓
Subtrahiere und vertausche die Seiten.
$0.10 x$ $=$ $5$ $: 0.1$
$x$ $=$ $\frac{5}{0.1}$
↓
Dividiere.
$x_{S}$ $=$ $50$
$x_{S} = 50$
Plug x into one of the two function equations.
$y$ $=$ $0.05 \cdot 50 + 20$
↓
Multipliziere.
$y$ $=$ $2.5 + 20$
↓
Addiere.
$y_{S}$ $=$ $22.5$
$\Rightarrow S (x_{S}; y_{S}) = S (50; 22.5)$
$S$ is the intersection point.
Sketch
Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ to obtain the lines.

Calculate the intersection points of a straight line.

Consider are the function equations of the two lines $g_{1} (x)$ and $g_{2} (x)$ . Calculate the intersection of the two lines and draw the lines in a coordinate system.

$g_{1} (x) = \frac{1}{2} x + 2 g_{2} (x) = - \frac{1}{2} x + 4$

For this task you need the following basic knowledge: Linear function
Calculating the line intersection
Set $g_{1} (x)$ and $g_{2} (x)$ equal.
$\frac{1}{2} x + 2$ $=$ $- \frac{1}{2} x + 4$ $| + \frac{1}{2} x - 2$
$\frac{1}{2} x + \frac{1}{2} x$ $=$ $4 - 2$
$x_{S}$ $=$ $2$
↓
Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$ $=$ $1 + 2$
$y_{S}$ $=$ $3$
$\Rightarrow S (x_{S} | y_{S}) = S (2 | 3)$
Sketch
Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.
$g_{1} (x) = 2 x - 1 g_{2} (x) = - 2 x + 1$

For this task you need the following basic knowledge: Linear function
Calculating the line intersection
Set $g_{1} (x)$ and $g_{2} (x)$ equal.
$2 x - 1$ $=$ $- 2 x + 1$ $+ 2 x + 1$
$2 x + 2 x$ $=$ $1 + 1$
$4 x$ $=$ $2$ $: 4$
$x_{S}$ $=$ $\frac{1}{2}$
↓
Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$ $=$ $2 \cdot \frac{1}{2} - 1$
$y$ $=$ $1 - 1$
$y_{S}$ $=$ $0$
$\Rightarrow S (x_{S} | y_{S}) = S (\frac{1}{2} | 0)$
Sketch
Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.
$g_{1} (x) = \frac{3}{4} x - 4 g_{2} (x) = - \frac{1}{2} x - 1$

For this task you need the following basic knowledge: Linear function
Calculating the line intersection
Set $g_{1} (x)$ and $g_{2} (x)$ equal.
$g_{1} (x)$ $=$ $g_{2} (x)$
$\frac{3}{4} x - 4$ $=$ $- \frac{1}{2} x - 1$ $+ \frac{1}{2} x + 4$
$\frac{3}{4} x + \frac{1}{2} x$ $=$ $- 1 + 4$
$1.25 x$ $=$ $3$ $1.25$
$x$ $=$ $\frac{3}{1.25}$
$x_{S}$ $=$ $2.4$
↓
Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$ $=$ $\frac{3}{4} \cdot 2.4 - 4$
$y$ $=$ $1.8 - 4$
$y_{S}$ $=$ $- 2.2$
$\Rightarrow S (x_{S} | y_{S}) = S (2.4 | - 2.2)$
Sketch
Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.
$g_{1} (x) = - \frac{1}{2} x + 2 g_{2} (x) = \frac{1}{2} x + 3$

For this task you need the following basic knowledge: Linear function
Calculating the line intersection
Set $g_{1} (x)$ and $g_{2} (x)$ equal.
$g_{1} (x)$ $=$ $g_{2} (x)$
$- \frac{1}{2} x + 2$ $=$ $\frac{1}{2} x + 3$ $| + \frac{1}{2} x - 3$
$2 - 3$ $=$ $\frac{1}{2} x + \frac{1}{2} x$
$x_{S}$ $=$ $- 1$
↓
Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$ $=$ $- \frac{1}{2} \cdot (- 1) + 2$
$y$ $=$ $\frac{1}{2} + 2$
$y_{S}$ $=$ $2.5$
$\Rightarrow S (x_{S} | y_{S}) = S (- 1 | 2.5)$
Sketch
Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.

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

$g_{1} (x) = \frac{3}{4} x + 1 g_{2} (x) = \frac{1}{2} x + 2$

4
Consider the following graphs.
Determine the function equations of all 4 lines.

For this task you need the following basic knowledge: 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
$m_{f} = \frac{y_{B} - y_{A}}{x_{B} - x_{A}}$
Plug in the values
$m_{f} = \frac{2 - 3}{4 - 0} = - \frac{1}{4}$
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 $A$ .
$3 = - \frac{1}{4} \cdot 0 + b_{f}$
Simplify.
$3 = b_{f} \Rightarrow b_{f} = 3$
So the line equation is $f (x) = - \frac{1}{4} \cdot x + 3$ .
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
$m_{g} = \frac{y_{D} - y_{C}}{x_{D} - x_{C}}$
Plug in the values
$m_{g} = \frac{1 - 0}{0 - (- 4)} = \frac{1}{4}$
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 $D$ .
$1 = \frac{1}{4} \cdot 0 + b_{g}$
Simplify.
$1 = b_{g} \Rightarrow b_{g} = 1$
So the line equation is $g (x) = \frac{1}{4} \cdot x + 1$ .
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
$m_{h} = \frac{y_{A} - y_{E}}{x_{A} - x_{E}}$
Plug in the values
$m_{h} = \frac{3 - 0}{0 - (- 1)} = 3$
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 $A$ .
$3 = 3 \cdot 0 + b_{h}$
Simplify.
$3 = b_{h} \Rightarrow b_{h} = 3$
So the line equation is $h (x) = 3 \cdot x + 3$ .
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
$m_{i} = \frac{y_{S} - y_{F}}{x_{S} - x_{F}}$
Plug in the values
$m_{i} = \frac{0 - (- 3)}{6 - 0} = \frac{1}{2}$
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 $F$ .
$- 3 = \frac{1}{2} \cdot 0 + b_{i}$
Simplify.
$- 3 = b_{i} \Rightarrow b_{i} = - 3$
So the line equation is $i (x) = \frac{1}{2} \cdot x - 3$ .

Determine the intersection of g and h , and the zero of f.

Intersection point $P (x_{p} | y_{p})$ of $g$ and $h$
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 $g (x) : y = \frac{1}{4} x + 1$ and $h (x) : y = 3 x + 3$ .
$\frac{1}{4} x_{P} + 1$ $=$ $3 x_{P} + 3$ $- 3 x_{p} - 1$
↓
subtract $3 x_{P}$ and $1$ .
$- \frac{11}{4} x_{P}$ $=$ $2$ $\div (- \frac{11}{4})$
↓
divide by $- \frac{11}{4}$ .
$x_{p}$ $=$ $- \frac{8}{11}$
$=$
Now substitute $- \frac{8}{11}$ into the line equation of $g$ or $h$ to determine $y_{P}$ .
$h (x_{P}) : y_{P} = 3 \cdot x_{P} + 3$
Plug in $x_{P}$ .
$y_{P} = 3 \cdot (- \frac{8}{11}) + 3 = \frac{9}{11}$
The lines $g$ and $h$ therefore intersect at $P (- \frac{8}{11} | \frac{9}{11})$ .
You may obtain the zero $x_{N_{f}}$ of $f$ by setting the function equation $f (x) : y = - \frac{1}{4} x + 3$ equal to 0 and solving for $x$ .
$- \frac{1}{4} x_{N_{f}} + 3$ $=$ $0$ $- 3$
$- \frac{1}{4} x_{N_{f}}$ $=$ $- 3$ $: \frac{1}{4}$
$x_{N_{f}}$ $=$ $12$
So $f$ has the zero 12.

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 $i (x) : y = \frac{1}{2} x - 3$ .
$3 x_{T} + 3$ $=$ $\frac{1}{2} x_{T} - 3$ $- \frac{1}{2} x - 3$
$\frac{5}{2} x_{T}$ $=$ $- 6$ $: \frac{5}{2}$
$x_{T}$ $=$ $- \frac{12}{5}$
Now substitute $- \frac{12}{5}$ into the line equation of $h$ or $i$ to determine $y_{T}$ .
$h (x_{T}) : y_{T} = 3 \cdot x_{T} + 3$
Plug in $x_{T}$ .
$y_{T} = 3 \cdot (- \frac{12}{5}) + 3 = - \frac{21}{5}$
The lines $h$ and $i$ therefore intersect at $T (- \frac{5}{2} | - \frac{21}{5})$ .
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 $g (x) : y = \frac{1}{4} x + 1$ and $i (x) : y = \frac{1}{2} x - 3$ .
$\frac{1}{4} x_{Q} + 1$ $=$ $\frac{1}{2} x_{Q} - 3$ $- \frac{1}{2} x - 1$
$- \frac{1}{4} x_{Q}$ $=$ $- 4$ $: (- \frac{1}{4})$
$x_{Q}$ $=$ $16$
Now substitute $16$ into the line equation of $g$ or $i$ to determine $y_{Q}$ .
$g (x_{Q}) : y_{Q} = \frac{1}{4} \cdot x_{Q} + 1$
Plug in $x_{Q}$ .
$y_{Q} = \frac{1}{4} \cdot 16 + 1 = 5$
The lines $g$ and $i$ therefore intersect at $Q (16 | 5)$ .

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 $i$

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.
$7 x - 14$ $=$ $7 x - 3$ $- 7 x$
$- 14$ $=$ $- 3$
↓
False statement!

$y = \frac{1}{6} x - 4$ 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.
$x - 1.5$ $=$ $- 2 x + 7.5$
↓
First add 2x and then 1.5.
$3 x$ $=$ $9$
↓
divide by 3.
$x$ $=$ $3$
$=$
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.
$1.5 = 0.5 \cdot 3$
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.
7
Check whether the following lines $g, h$ and $i$ run through a common point.
$g (x) = x + 1; h : 2 y + x + 4 = 0; 3 y - 5 x = 7$

For this task you need the following basic knowledge: 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 $h$ :
$2 y + x + 4 = 0$
$\Leftrightarrow 2 y = - x - 4$
$\Leftrightarrow y = - \frac{1}{2} x - 2$
Line $i$ :
$3 y - 5 x = 7$
$\Leftrightarrow 3 y = 5 x + 7$
$\Leftrightarrow y = \frac{5}{3} x + \frac{7}{3}$
Determine the intersection of $g$ and $h$
Set the respective right sides equal.
$x + 1$ $=$ $- \frac{1}{2} x - 2$ $+ \frac{1}{2} x - 1$
↓
Get all $x$ -terms on one side.
$x + \frac{1}{2} x$ $=$ $- 2 - 1$
↓
Conclude
$\frac{3}{2} x$ $=$ $- 3$ $: \frac{3}{2}$
$x$ $=$ $- 2$
↓
Plug into $g$ to get the $y$ -coordinate.
$y$ $=$ $- 2 + 1 = - 1$
The intersection point is $S_{g h} (- 2 | - 1) .$
Determine the intersection of $g$ and $i$
Set the respective right sides equal.
$x + 1$ $=$ $\frac{5}{3} x + \frac{7}{3}$ $- \frac{5}{3} x - 1$
↓
Get all $x$ -terms on one side.
$x - \frac{5}{3} x$ $=$ $\frac{7}{3} - 1$
↓
Conclude.
$- \frac{2}{3} x$ $=$ $\frac{4}{3}$ $: (- \frac{2}{3})$
$x$ $=$ $- 2$
↓
Plug into $g$ to get the $y$ -coordinate.
$y$ $=$ $- 2 + 1 = - 1$
The intersection point is $S_{g i} (- 2 | - 1) .$
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 $(- 2 | - 1)$ .

$g (x) = \frac{1}{6} x + \frac{3}{2}; h (x) = - \frac{2}{3} x + 2; i : 2 x - y = 3$

For this task you need the following basic knowledge: Linear function
First bring all lines equations into the general form $y = m x + t$ .
Line $g$ :
$g (x) = \frac{1}{6} x + \frac{3}{2}$
$\Leftrightarrow y = \frac{1}{6} x + \frac{3}{2}$
Line $h$ :
$h (x) = - \frac{2}{3} x + 2$
$\Leftrightarrow y = - \frac{2}{3} x + 2$
Line $i$ :
$2 x - y = 3$
$\Leftrightarrow y = 2 x - 3$
Determine the intersection of $g$ and $h$
Set the respective right sides equal.
$\frac{1}{6} x + \frac{3}{2}$ $=$ $- \frac{2}{3} x + 2$ $+ \frac{2}{3} x - \frac{3}{2}$
↓
Get all $x$ -terms on one side.
$=$
↓
Conclude.
$\frac{1}{6} x + \frac{2}{3} x$ $=$ $2 - \frac{3}{2}$
$\frac{5}{6} x$ $=$ $\frac{1}{2}$ $: \frac{5}{6}$
$x$ $=$ $\frac{3}{5}$
↓
Plug into $g$ to get the $y$ -coordinate.
$y$ $=$ $\frac{1}{6} \cdot \frac{3}{5} + \frac{3}{2}$
$y$ $=$ $\frac{1}{10} + \frac{3}{2} = \frac{16}{10}$
$y$ $=$ $\frac{8}{5}$
The intersection point is $S_{g h} (\frac{3}{5} | \frac{8}{5}) .$
Determine the intersection of $g$ and $i$
Set the respective right sides equal.
$\frac{1}{6} x + \frac{3}{2}$ $=$ $2 x - 3$ $- 2 x - \frac{3}{2}$
↓
Get all $x$ -terms on one side.
$\frac{1}{6} x - 2 x$ $=$ $- 3 - \frac{3}{2}$
↓
Conclude.
$- \frac{11}{6} x$ $=$ $- \frac{9}{2}$ $: (- \frac{11}{6})$
$x$ $=$ $\frac{9}{2} \cdot \frac{6}{11} = \frac{27}{11}$
↓
Plug into $g$ to get the $y$ -coordinate.
$y$ $=$ $\frac{1}{6} \cdot \frac{27}{11} + \frac{3}{2}$
$y$ $=$ $\frac{9}{22} + \frac{3}{2} = \frac{42}{44}$
$y$ $=$ $\frac{21}{11}$
The intersection point is $S_{g i} (\frac{27}{11} | \frac{21}{11}) .$
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.
8
Determination of intersection points
Consider the following two lines $g$ and $h$ .
Determine the line equations of $g$ and $h$ .

For this task you need the following basic knowledge: Line equation
$y = m x + t$
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 $t_{h} = 1$ .
$y = 2 x + 1$

Read off the intersection point.

For this task you need the following basic knowledge: Lines in coordinate systems
Reading off the intersection point
From the plot, we may read off the intersection point $SP (1 | 3)$ .

$- 3 x + \frac{5}{4}$	$=$	$- x - 1$	$+ 3 x + 1$
$- x + 3 x$	$=$	$\frac{5}{4} + 1$
	↓	add
$2 x$	$=$	$\frac{9}{4}$	$: 2$
$x$	$=$	$\frac{9}{8}$

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

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

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

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

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

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

$y$	$=$	$0.05 \cdot 50 + 20$
	↓	Multipliziere.
$y$	$=$	$2.5 + 20$
	↓	Addiere.
$y_{S}$	$=$	$22.5$

$\frac{1}{2} x + 2$	$=$	$- \frac{1}{2} x + 4$	$\| + \frac{1}{2} x - 2$
$\frac{1}{2} x + \frac{1}{2} x$	$=$	$4 - 2$
$x_{S}$	$=$	$2$
	↓	Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$	$=$	$1 + 2$
$y_{S}$	$=$	$3$

$2 x - 1$	$=$	$- 2 x + 1$	$+ 2 x + 1$
$2 x + 2 x$	$=$	$1 + 1$
$4 x$	$=$	$2$	$: 4$
$x_{S}$	$=$	$\frac{1}{2}$
	↓	Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$	$=$	$2 \cdot \frac{1}{2} - 1$
$y$	$=$	$1 - 1$
$y_{S}$	$=$	$0$

$\frac{1}{4} x_{P} + 1$	$=$	$3 x_{P} + 3$	$- 3 x_{p} - 1$
	↓	subtract $3 x_{P}$ and $1$ .
$- \frac{11}{4} x_{P}$	$=$	$2$	$\div (- \frac{11}{4})$
	↓	divide by $- \frac{11}{4}$ .
$x_{p}$	$=$	$- \frac{8}{11}$
	$=$

$- \frac{1}{4} x_{N_{f}} + 3$	$=$	$0$	$- 3$
$- \frac{1}{4} x_{N_{f}}$	$=$	$- 3$	$: \frac{1}{4}$
$x_{N_{f}}$	$=$	$12$

$3 x_{T} + 3$	$=$	$\frac{1}{2} x_{T} - 3$	$- \frac{1}{2} x - 3$
$\frac{5}{2} x_{T}$	$=$	$- 6$	$: \frac{5}{2}$
$x_{T}$	$=$	$- \frac{12}{5}$

$\frac{1}{4} x_{Q} + 1$	$=$	$\frac{1}{2} x_{Q} - 3$	$- \frac{1}{2} x - 1$
$- \frac{1}{4} x_{Q}$	$=$	$- 4$	$: (- \frac{1}{4})$
$x_{Q}$	$=$	$16$

$g_{1} (x)$	$=$	$g_{2} (x)$
$\frac{3}{4} x - 4$	$=$	$- \frac{1}{2} x - 1$	$+ \frac{1}{2} x + 4$
$\frac{3}{4} x + \frac{1}{2} x$	$=$	$- 1 + 4$
$1.25 x$	$=$	$3$	$1.25$
$x$	$=$	$\frac{3}{1.25}$
$x_{S}$	$=$	$2.4$
	↓	Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$	$=$	$\frac{3}{4} \cdot 2.4 - 4$
$y$	$=$	$1.8 - 4$
$y_{S}$	$=$	$- 2.2$

$g_{1} (x)$	$=$	$g_{2} (x)$
$- \frac{1}{2} x + 2$	$=$	$\frac{1}{2} x + 3$	$\| + \frac{1}{2} x - 3$
$2 - 3$	$=$	$\frac{1}{2} x + \frac{1}{2} x$
$x_{S}$	$=$	$- 1$
	↓	Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$	$=$	$- \frac{1}{2} \cdot (- 1) + 2$
$y$	$=$	$\frac{1}{2} + 2$
$y_{S}$	$=$	$2.5$

$g_{1} (x)$	$=$	$g_{2} (x)$
$\frac{2}{3} x + 2$	$=$	$\frac{1}{2} x + 3$	$- \frac{1}{2} x - 2$
	↓	Get the fractions on a common denominator.
$\frac{2}{3} x - \frac{1}{2} x$	$=$	$3 - 2$
	↓	subtract
$\frac{4}{6} x - \frac{3}{6} x$	$=$	$3 - 2$
$\frac{1}{6} x$	$=$	$1$	$: \frac{1}{6}$
	↓	dividing by a fraction $\to$ multiplying with the inverse
$x$	$=$	$\frac{1}{\frac{1}{6}}$
$x_{S}$	$=$	$6$
	↓	Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$	$=$	$\frac{2}{3} \cdot 6 + 2$
	↓	add
$y$	$=$	$4 + 2$
$y_{S}$	$=$	$6$

$g_{1} (x)$	$=$	$g_{2} (x)$
$\frac{3}{4} x + 1$	$=$	$\frac{1}{2} x + 2$	$- \frac{1}{2} x - 1$
$\frac{3}{4} x - \frac{1}{2} x$	$=$	$2 - 1$
	↓	Get the fractions on a common denominator.
$\frac{3}{4} x - \frac{2}{4} x$	$=$	$2 - 1$
	↓	subtract
$\frac{1}{4} x$	$=$	$1$	$: \frac{1}{4}$
	↓	dividing by a fraction $\to$ multiplying with the inverse
$x$	$=$	$\frac{1}{\frac{1}{4}}$
$x_{S}$	$=$	$4$
	↓	Plug $x = x_{s}$ into $g_{1} (x)$ .
$y$	$=$	$\frac{3}{4} \cdot 4 + 1$
	↓	shorten
$y$	$=$	$3 + 1$
	↓	add
$y_{S}$	$=$	$4$

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

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

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

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

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

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

Exercises: Intersection of two lines

Sketch

Determining the intersection point

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Calculating the line intersection

Sketch

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

Sketch

Calculating the line intersection

Sketch

Line equation of f

Line equation of g

Line equation of h

Line equation of i

Intersection point P(xp|yp) of g and h

Intersection point T(xT|yT) of h and i

Intersection point Q(xQ|yQ) 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

Determination of intersection points

Reading off the intersection point

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 $i$

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

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

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

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

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