What is the point-slope equation of the line with a slope 4/3 that goes through the point (-4,6)?

[tex]\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{6})~\hspace{10em}slope = m\implies \cfrac{4}{3}\\\\\\\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-6=\cfrac{4}{3}[x-(-4)]\implies y-6=\cfrac{4}{3}(x+4)[/tex]
Answer:
The point-slope form of the equation is y - 6 = 4/3(x + 4), which is answer B.
Step-by-step explanation:
In order to find the point-slope form of any equation, start with the base form of the equation.
y - y1 = m(x - x1)
Now input the slope given as m, and the point in for x1 and y1.
y - y1 = m(x - x1)
y - 6 = 4/3(x - -4)
y - 6 = 4/3(x + 4)