Posted by **DJ Frankie Pigeon** on Monday, January 15, 2007 at 3:09pm.

1. Find the equation of the line described in each case:

a) Gradient 3, cuts through (0, 6).

b) Steepness of 4, passes through (0, -2).

c)Parallel to y = 5x - 1, cuts the y-axis at (0, 10).

d)Cuts through (0, 7) and (5, 17)

2. Here are 3 eqautions:

A. y + 1 = 3x

B. y = -3(x - 1)

C. 2y = 8x + 6

Find the odd one out in each case:

a) Gradient is ±3

b) Slopes in a positive direction

c) Cuts the y-axis at (0, 3)

3. A line passes through the point (10, 10). It's equation is y = mx + 4. Find the value of m.

I'll try to get you started on these, and let you take it from there.

For 1a:

y - y1 = m(x - x1)

y - 6 = 3(x - 0)

y - 6 = 3x

y = 3x + 6

For 1b:

Use the same procedure as 1a.

For 1c:

Parallel lines have the same slope.

y - 10 = 5(x - 0)

I'll let you finish.

For 1d:

m = (y2 - y1)/(x2 - x1)

m = (17 - 7)/(5 - 0)

m = 10/5 = 2

Therefore, using the first ordered pair (0, 7), we have this:

(y - 7)/(x - 0) = 2

y - 7 = 2(x - 0)

I'll let you take it from here to finish.

For 2a:

The coefficient of x is the slope.

For 2b:

Look for positive slopes.

For 2c:

Substitute 0 for x and 3 for y in each equation to check each one.

For 3:

Substitute 10 for x and 10 for y. Solve the equation for m, which is the slope.

I hope this will help get you started on these problems.

<<3. A line passes through the point (10, 10). It's equation is y = mx + 4. Find the value of m>>

Since y = mx + 4 and y = 10 when x = 10, plug those values of x and y into the eqwuation and solve for m.

Make an attempt at the others and we will glad to critique your work. Please limit your questions to one per post.

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