Answer:
[tex]\dfrac{27}{14}\ un^2.[/tex]
Step-by-step explanation:
Plot the graphs of the given lines on hte coordinate plane (see attached diagram).
These three lines form triangle ABC.
1. Find the length of AC:
Point A has coordinates (0,0), point C has coordinates (0,3) {point of intersection of the line [tex]y=-\frac{4}{3}x+3[/tex] with y-axis}. Hence,
[tex]AC=|3-0|=3\ units[/tex]
2. Find the distance from point B to the y-axis:
First, find the coordinates of the point B:
[tex]y=x\\ \\y=-\dfrac{4}{3}x+3\Rightarrow x=-\dfrac{4}{3}x+3,\ \ \dfrac{7}{3}x=3,\ \ x=y=\dfrac{9}{7}[/tex]
And point B has coordinates [tex]\left(\dfrac{9}{7},\dfrac{9}{7}\right)[/tex]
So,
[tex]h=\dfrac{9}{7}\ units[/tex]
3. Find the area of the triangle ABC:
[tex]A_{ABC}=\dfrac{1}{2}AC\cdot h=\dfrac{1}{2}\cdot 3\cdot \dfrac{9}{7}=\dfrac{27}{14}\ un^2.[/tex]
