Exercises: Linear functions as lines in coordinate systems

Area of the triangle

Since the coordinate axes are perpendicular to each other, they can be used as the baseline and height of the formula for calculating the area of ​​the triangle.

Determine the height

g: $y=\frac{2}{3}x+5$

The y-intercept is given as t of the line equation.

The straight line intersects the y-axis in (0|5).

With the x-axis as the baseline, the distance from the point to the origin is the height.

$\Rightarrow$ The height h =$\sqrt{5^2-0^2}=5$

Calculate baseline

To specify the baseline, the point of intersection of the straight line with the x-axis must be calculated. Set y to 0 for this.

$0=\frac{2}{3}x+5$

Solve for x. $|-5$

$\frac{2}{3}x=-5$

$|:\frac{2}{3}$

$x=-\frac{15}{2}=-7.5$

The straight line intersects the x-axis in (-7,5|0)

So the distance from this point to the origin is the length of the baseline.

$\Rightarrow$ The length of the baseline g is $\sqrt{(-7.5{)}^2+0^2}=7.5$.

Calculate the area

Substitute the baseline and height into the formula to calculate the area of ​​the triangle.

$\Rightarrow A=\frac{5\cdot \mathrm{7,5}}{2}=18.75\mathrm{AU}$

AU stays for "area unit".

Illustration: