Question

Please help me with these two questions, 10 points for each so in total 20!!

Answer 1

Answer:

$\large\boxed{Q2.\qquad C.\ -2x+y=-2}\\\boxed{Q4.\qquad C.\ y=3x+12}$

Step-by-step explanation:

Q2:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (4, 6) and (6, 10). Substitute:

$m=\dfrac{10-6}{6-4}=\dfrac{4}{2}=2$

$y-6=2(x-4)$           use distributive property

$y-6=2x-8$      add 6 to both sides

$y=2x-2$          subteact 2 from both sides

$-2x+y=-2$

Q4:

The slope-intercept form of an equation of a line:

$y=mx+b$

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:

$y-6=3(x-(-2))$

$y-6=3(x+2)$         use distributive property

$y-6=3x+6$     add 6 to both sides

$y=3x+12$