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

2- Which is an equation of the line containing the points (4,6) and (6,10) in standard form?

A: -2x+y = -8

B: -2-y = -22

C: -2+y= -2

D: 2x-y= -2

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4- Which shows the equation of the line containing the point (-2,6) And having a slope of 3 in-intercept form?

A: 3y=x+20

B: y=3x-20

C: y=3x+12

D: y= -3/2x+3

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gmany gmany · Answer 1 · 2019-04-17T22:49:00+02:00

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]