Question

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

Answer 1

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

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=(y_2-y_1)/(x_2-x_1)`

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

`m=(10-6)/(6-4)=(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`