The equation y=x-2 do not represents the line that passes through the two point, since for example -4-2=-6, not 10. So (-4,10) is not on the line.
Let find the equation of the line:
First compute the slope using formula:
[tex] \frac{y_2-y_1}{x_2-x_1}= \frac{5-10}{-1-(-4)}= \frac{-5}{3} [/tex]
So the equation is in the form:
[tex]y=\frac{-5}{3}x+s\text{ wherein s is another parameter. }[/tex]
For x=-4 we get the equation:
[tex]10=\frac{-5}{3}(-4)+s[/tex]
Solving the above equation for s we get:
[tex]s= \frac{10}{3} [/tex]
The equation then is the following:
[tex]y=\frac{-5}{3}x+\frac{10}{3}[/tex]
The correct answer then is equation 2 and equation 3.
