Exercises: Intersection of two 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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