Explanation
when you have the slope and a point of the line, you can find the equation by using Point-slope is the general form
[tex]y-y_{1}=m\mleft(x-x_{1}\mright)[/tex]where
[tex]\begin{gathered} \text{m is the slope and} \\ (x_1,y_1)\text{ is a point of the line} \end{gathered}[/tex]
so
Step 1
Let
slope=-4/3
point=(-4,2)
replace and isolate y
[tex]\begin{gathered} y-y_1=m\mleft(x-x_1\mright) \\ y-2=\frac{4}{3}(x-(-4)) \\ y-2=\frac{4}{3}x+\frac{16}{3} \\ ad\text{d 2 in both sides} \\ y-2+2=\frac{4}{3}x+\frac{16}{3}+2 \\ y=\frac{4}{3}x+\frac{22}{3} \end{gathered}[/tex]so, the equation of the line is
[tex]y=\frac{4}{3}x+\frac{22}{3}[/tex]Step 2
plot 3 points
to do that, replace in the equation and you will get the y -coordinate
so
a) when x=0
[tex]\begin{gathered} y=\frac{4}{3}\cdot0+\frac{22}{3} \\ y=0+\frac{22}{3} \\ y=\frac{22}{3} \\ P(0,\frac{22}{3}) \end{gathered}[/tex]b) when x= 1
[tex]\begin{gathered} y=\frac{4}{3}\cdot1+\frac{22}{3} \\ y=\frac{4}{3}+\frac{22}{3} \\ y=\frac{26}{3} \\ P(1,\frac{26}{3}) \end{gathered}[/tex]c) when x= -1
[tex]\begin{gathered} y=\frac{4}{3}\cdot-1+\frac{22}{3} \\ y=-\frac{4}{3}+\frac{22}{3} \\ y=\frac{18}{3}=6 \\ P(-1,6) \end{gathered}[/tex]Step 3
finally, draw a line that passes through the points
I hope this helps you
