SOLUTION: Create multiple representation of each line described below. A) A line with a slope of 4 and y-intercept-6. B) A line with a slope of 3/2 that passes through the point (5,7). C)

Algebra ->  Linear-equations -> SOLUTION: Create multiple representation of each line described below. A) A line with a slope of 4 and y-intercept-6. B) A line with a slope of 3/2 that passes through the point (5,7). C)      Log On


   



Question 1133741: Create multiple representation of each line described below.
A) A line with a slope of 4 and y-intercept-6.
B) A line with a slope of 3/2 that passes through the point (5,7).
C)A line parallel to the line in part(b).
D) A line perpendicular to the line in part (b), passing through the origin.
Thank you so much.

Answer by Wize(1) About Me  (Show Source):
You can put this solution on YOUR website!
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+=+%28-2%2F3%29x
or
y+%2B+%282%2F3%29x+=+0+
Hope you get this help and can apply it more!