A couple quiz questions please help!?

1-6. These are all done the same way, so I'll do a couple and you should be able to do the remainder.

You're given a point on the line and the slope of the line. So use the point-slope format of the equation of the line, and then convert it into the slope-intercept format.

y - y1 = m(x - x1) <----- point-slope format

y - 6 = 2(x - 2) <--------- fill in m, x1, & y1, where (x1,y1) is the point on the line.

y - 6 = 2x - 4

y = 2x + 2 <--------------- slope-intercept format

y - y1 = m(x - x1)

y - (-9) = -5(x - 3)

y + 9 = -5x + 15

y = -5x + 6

-- etc --

8. Perpendicular lines have slopes that are the negative reciprocol of each other. So for the line y = -2x + 6, the slope is -2, and all lines perpendicular to it will have slope 1/2 (the negative reciprocol of -2). So now you know the slope of the perpendicular line and a point on that line, so do the same thing I did above. The difference, though, is that this problem asks for the final answer in standard form instead of slope-intercept form.

y - y1 = m(x - x1)

y - 7 = (1/2)(x - (-4))

y - 7 = (1/2)(x + 4)

2(y - 7) = (x + 4) <----- Multiply each side by 2 to remove the denominator.

2y - 14 = x + 4

0 = x - 2y + 18 <------- Equation in standard format.

9-15) As with the first set of problems, I'll do a couple and you should be able to do the remainder. Standard form is ax + by + c = 0, where a, b, and c are integer coefficients AND a > 0.

Step 1: If there are any fractions in the equation, multiply each side by the LCM to remove them.

Step 2: Rearrange the terms into Standard Format.

That's all there is to it.

1/2 - x = 9y

2(1/2 - x) = 18y

1 - 2x = 18y

0 = 2x + 18y - 1

y = (5/13)x + 4

13y = 13((5/13)x + 4)

13y = 5x + 52

0 = 5x - 13y + 52

6y = -18x + 3

18x + 6y - 3 = 0