Answer:
[tex]\textsf{b.} \quad 3x+5y=2[/tex]
Step-by-step explanation:
Define the given points:
[tex]\textsf{let}\:(x_1,y_1)=(4,-2)[/tex]
[tex]\textsf{let}\:(x_2,y_2)=(-1,1)[/tex]
Use the slope formula to determine the slope of the line:
[tex]\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{1-(-2)}{-1-4}=\dfrac{1+2}{-1-4}=-\dfrac{3}{5}[/tex]
Use the point-slope formula to determine the equation of the line.
[tex]\implies y-y_1=m(x-x_1)[/tex]
Substitute the found slope and point (4, -2):
[tex]\implies y-(-2)=-\dfrac{3}{5}(x-4)[/tex]
[tex]\implies y+2=-\dfrac{3}{5}(x-4)[/tex]
Multiply both sides by 5:
[tex]\implies 5(y+2)=-3(x-4)[/tex]
Expand the brackets:
[tex]\implies 5y+10=-3x+12[/tex]
Add 3x to both sides:
[tex]\implies 5y+10+3x=-3x+12+3x[/tex]
[tex]\implies 3x+5y+10=12[/tex]
Subtract 10 from both sides:
[tex]\implies 3x+5y+10-10=12-10[/tex]
[tex]\implies 3x+5y=2[/tex]
