The slope intercept form of the line is given by:
[tex]y = mx + b [/tex]
The point-slope form is given by:
[tex]y-yo = m (x-xo) [/tex]
Where,
m: slope of the line
b: cutting point with the y axis
(xo, yo): ordered pair that belongs to the line
For line A:
The slope is
[tex]m = \frac{1-3}{4-1} [/tex]
[tex]m = \frac{-2}{3} [/tex]
The cut point with the y axis is:
[tex]b = \frac{11}{3} [/tex]
Substituting values we have:
[tex]y = -\frac{2}{3}x + \frac{11}{3} [/tex]
For line B:
The slope is given by:
[tex]m=\frac{1-(-5)}{4-0}[/tex]
[tex]m=\frac{1+5}{4}[/tex]
[tex]m=\frac{6}{4}[/tex]
[tex]m=\frac{3}{2}[/tex]
Then, the equation of the line is:
[tex]y-1= \frac{3}{2}(x-4) [/tex]
Answer:
Option A
