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Exercises: Linear functions as lines in coordinate systems

The straight lines are given $g$ : $y = 2 x - 3$ and $h$ : $y = - 0.5 x + 4$ .

Calculate the intersection of the two lines.

g: $y = 2 x - 3$ and h: $y = - 0.5 x + 4$
Equate the lines to get the x-value of the intersection.
$2 x - 3$ $=$ $- 0.5 x + 4$ $+ 0.5 x | + 3$
$2.5 x$ $=$ $7$ $: 2.5$
$x$ $=$ $2.8$
$y$ $=$ $2 \cdot 2.8 - 3$
↓
Note "dot before dash"
$y$ $=$ $5.6 - 3 = 2.6$
The x-value must now be inserted into one of the lines to get the y-value of the point of intersection.
$y$ $=$ $2 \cdot 2.8 - 3$
$y$ $=$ $5.6 - 3$
$y$ $=$ $2.6$
The intersection of the lines: $S (2.8 | 2.6)$
Calculate the area of the triangle formed by g and h and the y-axis.

The base line g of the triangle is 7
Here, starting from the formula for calculating the area of a triangle, the section between the two intersection points of the straight lines with the y-axis is taken as the base line. These intersections are marked with the letter t. For the given straight lines, the range lies between $- 3$ and $4$ $\Rightarrow$ $g = 7$
The height h is 2.8. The height h is always perpendicular to the base line and is therefore parallel to the x-axis. It thus corresponds to the length from the origin to the x-value of the intersection of both straight lines.
$\Rightarrow A = \frac{1}{2} \cdot g \cdot h = \frac{7 \cdot 2.8}{2} = 9.8$

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