[tex]\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{3})~\hspace{10em} slope = m\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-3=-4[x-(-2)]\implies y-3=-4(x+2)[/tex]
Answer: [tex](y-3)=(-4)(x+2)[/tex]
Step-by-step explanation:
We know that the equation of a line in point-slope form that is passing through a point (a,b) and has slope m is given by :-
[tex](y-b)=m(x-a)[/tex]
Then, the point-slope form of a line with slope -4 that contains the point (-2,3) :-
[tex](y-3)=(-4)(x-(-2))\\\\\Rightarrow\ (y-3)=(-4)(x+2)[/tex]
Hence, the point-slope form of a line with slope -4 that contains the point (-2,3) is [tex](y-3)=(-4)(x+2)[/tex]
