a) Slope intercept form
[tex]y=mx+b[/tex]We see in the graph that the y-intercept is y=4, so b=4.
Then, using another known point like (3,0) we can calculate the slope:
[tex]\begin{gathered} y=mx+4 \\ 0=m\cdot3+4 \\ 0=3m+4 \\ 3m=-4 \\ m=-\frac{4}{3} \end{gathered}[/tex]Then, the equation becomes:
[tex]y=-\frac{4}{3}x+4[/tex]b) Point slope form
[tex]y-y_1=m(x-x_1)[/tex]We will use the known slope m=-4/3 and one of the points (3,0):
[tex]y-0=-\frac{4}{3}(x-3)[/tex]c) Perpendicular line that pases through the point (5,-4):
In order to be perpendicular, the line slopes has to be opposite reciprocals:
[tex]m_1=-\frac{1}{m_2}=-\frac{1}{(-\frac{4}{3})}=\frac{1}{\frac{4}{3}}=\frac{3}{4}[/tex]Then, the perpendicular line has a slope of 3/4.
To make it go through the point (5,-4), we replace this values in the equation and calculate the y-intercept b:
[tex]\begin{gathered} y=\frac{3}{4}x+b \\ -4=\frac{3}{4}\cdot5+b \\ -4=\frac{15}{4}+b \\ b=-4-\frac{15}{4}=-\frac{16}{4}-\frac{15}{4}=-\frac{31}{4} \end{gathered}[/tex]The equation of the perpendicular line that goes through (5,-4) is:
[tex]y=\frac{3}{4}x-\frac{31}{4}[/tex]d) Parallel line that goes through (0,3).
Parallel lines have the same slope.
In this case, the slope is -4/3.
If we replace the values of x and y with the point (0,3) we get thte y-intercept b:
[tex]\begin{gathered} y=-\frac{4}{3}x+b \\ 3=-\frac{4}{3}\cdot0+b \\ b=3 \end{gathered}[/tex]We could have already now that the intercept was 3 because this is the value when x=0.
The equation of the line is:
[tex]y=-\frac{4}{3}x+3[/tex]