Write an equation in standard form of the line that passes through the given point and has the given slope. 1. (-3, 2); m = 1 2. (0, 5); m = -2 3. (4, 0); m = 5

In the equation, A, B, and C are all constants and A is positive, while x and y are just left as such.

To find this equation however, from a given point and a given slope, you must first create another equation in a different form, and change it to that form. There is no way to find standard form, as of now, directly without using another equation (that I'm aware of).

Point slope form: y - = m (x - )

The y and x are left as such, m = slope, and & are the coordinates of any given point on the line.

Things needed for this equation: slope and one point on the line

Slope Intercept form: y = mx + b

The m is the slope, and the b is the y intercept. The y and x are left as such. You can calculate b from a point on the line, but it's harder than just putting it in point slope form.

Things needed: slope, y intercept

As we can see, the form that would be the easiest to use is point-slope as it only needs a given point and slope, and we are provided that.

Here's the equation again: y - = m (x - )

1. Given point: (-3, 2) Slope: 1

y - 2 = (x + 3)

2. Given point: (0, 5) Slope: -2

y - 5 = -2x

3. Given point: (4, 0) Slope: 5

y = 5 (x - 4)

As you have it in point slope form, the conversion is the last step. Distribute the slope to the values inside the parenthesis, and then multiply so all fractions become integers, and move the x and y to one side and the remaining number to the other.

1. y - 2 = x + 3

2. y - 5 = -2x

3. y = 5 (x - 4)

y = 5x - 20

-5x + y = -20

A must be positive, so divide everything by -1.

Hope this helped!

i would be happy to . most people on would also love to i'm sure. if you just want someone who has done algebra before and want a little more personal relationship than just begging strangers for , i can certainly be of service. i do have college myself so i might not be 24/7.