Question 1133741
For the lines, there are different general formulas to use to describe a line's path on the coordinate plane. One very common one is slope-intercept form:

y = mx + b

(m) is the slope to know how the line should keep going.
(b) is the y-intercept value to know what y-value to start at when at 0.

x and y represent all values on the plane that solve the equation when put together. You can't have the equation if the y-value doesn't equal what it should after plugging in the x-value and vice versa. 

So for problem A you probably need at least two forms of the described line. One will be in point-slope form and the other will be in standard as they are the two easiest. 

Standard form is generally written in the form:

Ax + By = C

A is a value that works together with the B value to describe how a line behaves with constant C. In reference to point-slope formula:

m = -A/B
A = -B(m)
B = -A/(m)

So, back to the problem, a line with a slope of 4 and y-intercept of -6 can easily be expressed as:


y = 4x - 6
by plugging in given values to our known form of equation

or by moving values over to another form:

6 = 4x - y

For part B, start the same way by plugging in the givens into the structured form (get b by multiplying m by a known x value, 5, and solving for the missing value needed to add to get the known y): 

y = (3/2)x - 1/2

or 

(3/2)x - y = 1/2


For part C, find a line parallel by using the same slope but changing the y-intercept. That way they move the same, but start at different points. 

So one easy example is:

y = (3/2)x + 1

or 

y - (3/2)x = 1


Lastly for part D (sorry just discovered formulas midway lol), a lane perpendicular to another has a slope that is a negative reciprocal. So, the new slope would be (-2/3) and if it passed the origin it would have a y-intercept of 0 (given (0,0)). 

{{{y = (-2/3)x}}}

or 

{{{y + (2/3)x = 0 }}}

Hope you get this help and can apply it more!