Answer:
[tex]\large\boxed{Q2.\qquad C.\ -2x+y=-2}\\\boxed{Q4.\qquad C.\ y=3x+12}[/tex]
Step-by-step explanation:
Q2:
The point-slope form of an equation of a line:
[tex]y-y_1=m(x-x_1)[/tex]
m - slope
The formula of a slope:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
We have the points (4, 6) and (6, 10). Substitute:
[tex]m=\dfrac{10-6}{6-4}=\dfrac{4}{2}=2[/tex]
[tex]y-6=2(x-4)[/tex] use distributive property
[tex]y-6=2x-8[/tex] add 6 to both sides
[tex]y=2x-2[/tex] subteact 2 from both sides
[tex]-2x+y=-2[/tex]
Q4:
The slope-intercept form of an equation of a line:
[tex]y=mx+b[/tex]
m - slope
b - y-intercept
Put the slope m = 3 and the coordinateso f the point (-2, 6) to the point-slope form of an equation of a line:
[tex]y-6=3(x-(-2))[/tex]
[tex]y-6=3(x+2)[/tex] use distributive property
[tex]y-6=3x+6[/tex] add 6 to both sides
[tex]y=3x+12[/tex]
