1. Write the equation of the line which has y-intercept (0, 5) and is perpendicular to the line with equation y = –3x + 1.

You're given the slope of a line y=-3x +1, is obviously -3. Therefore the perpindicular line you are trying to find has slope 1/3. Now you have the slope and a point on the line, use the line formula y-y1=m(x-x1) and rearrange.

y-5=(1/3)(x-0)

y-5=(1/3)x

y=(1/3)x + 5

2. Write the equation of the line that passes through point (–7, 9) with a slope of –2

Simpler than above, just use the line formula;

y-y1=m(x-x1)

y-9=-2(x+7)

y-9=-2x + 7

y=-2x +7+9

y=-2x +16

3. Find the slope and y-intercept.

–2x + 10y = –20

rearrange for and make y the subject, the coefficient of x is the slope and the added constant is the y-coordinate of the y-intercept;

-2x+10y=-20

10y=2x-20

y=(1/5)x -4

therefore the slope is 1/5 and the y intercept is (0,-4)

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remember the key concepts here; the relationship of slopes between perpindicular and parallel lines, how to use the line and slope formulas; and recognising the form y=mx+b in finding slopes and y-intercepts.