Exercises: Intersection of two lines

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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.

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function

Calculating the line intersection

Set $g_{1} (x)$ and $g_{2} (x)$ equal.

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

$\Rightarrow\;\;S\left(x_S|y_S\right)=S\left(2|3\right)$

Sketch

Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function

Calculating the line intersection

Set $g_{1} (x)$ and $g_{2} (x)$ equal.

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

$\Rightarrow\;\;S\left(x_S|y_S\right)=S\left(\frac12|0\right)$

Sketch

Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.

Do you have a question?

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function

Calculating the line intersection

Set $g_{1} (x)$ and $g_{2} (x)$ equal.

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

$\Rightarrow\;\;S\left(x_S|y_S\right)=S\left(2.4|-2.2\right)$

Sketch

Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.

Do you have a question?

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function

Calculating the line intersection

Set $g_{1} (x)$ and $g_{2} (x)$ equal.

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

$\Rightarrow\;\;S(x_S|y_S)=S\left(-1|2.5\right)$

Sketch

Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.

Do you have a question?

Write it here 👇

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function

Calculating the line intersection

Set $g_{1} (x)$ and $g_{2} (x)$ equal.

$\displaystyle g_1(x)$	$=$	$\displaystyle g_2(x)$
$\displaystyle \frac23x+2$	$=$	$\displaystyle \frac12x+3$	$\displaystyle -\frac{1}{2}x-2$
	↓	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
$\displaystyle y$	$=$	$\displaystyle 4+2$
$\displaystyle y_S$	$=$	$\displaystyle 6$

$\Rightarrow\;\;S(x_S|y_S)=S\left(6|6\right)$

Sketch

Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.

Do you have a question?

Write it here 👇

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

Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function

Calculating the line intersection

Set $g_{1} (x)$ and $g_{2} (x)$ equal.

$\displaystyle g_1(x)$	$=$	$\displaystyle g_2(x)$
$\displaystyle \frac{3}{4}x+1$	$=$	$\displaystyle \frac{1}{2}x+2$	$\displaystyle -\frac{1}{2}x-1$
$\displaystyle \frac{3}{4}x-\frac{1}{2}x$	$=$	$\displaystyle 2-1$
	↓	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
$\displaystyle y_S$	$=$	$\displaystyle 4$

$\Rightarrow\;\;S(x_S|y_S)=S\left(4|4\right)$

Sketch

Connect the $y$ -axis intercepts (here $A$ and $B$ ) with the calculated intersection point $S$ . This way you obtain the two lines.

Do you have a question?

Write it here 👇

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