Answer:
The "point-slope" form of the equation of a straight line is:
\begin{gathered}y-y_1=m(x-x_1)\\\\(x_1;\ y_1)\ is\ a\ known\ point\\\\m\ is\ the\ slope\ of\ the\ line\end{gathered}y−y1=m(x−x1)(x1; y1) is a known pointm is the slope of the line
We have:
m=-3;\ (x_1;\ y_1)=(2;-2)m=−3; (x1; y1)=(2;−2)
Substitute
\begin{gathered}y-(-2)=-3(x-2)\\\\y+2=-3(x-2)\end{gathered}y−(−2)=−3(x−2)y+2=−3(x−2)
Answer: B. y + 2 = -3(x - 2
