State whether the graphs of the following equations are parallel, perpenicular, or neither.?

Question

1. x + y = 5
x - y = 5

2. 2y + 3x = 5
3y - 2x = 5

3. 1/2x + 1/3y = 2

3.) 1/2x + 1/3y = 2
2x - 3y = 4

Thank you guys, because I don't know how to do this

Answer 1

I am not sure why you aren't just graphing this out and doing them yourself.

Good Luck to you!!!

Answer 2

The Answer is below an dthe solution is even below it:
1)Parallel
2)Perpendicular
3)Neither




Rule of thumb: In the subject of equations of lines,
parallel = "same slope"
perpendicular = "negative reciprocal"

That's exactly how we solve these; we find the slope m, and compare them. Note that negative reciprocal means the reciprocal of the slope in question, multiplied by -1.

1. x + y = 5, x - y = 5

The first step would be to convert your lines to slope-intercept for; this is of the form y = mx + b.

y = -x + 5, y = x - 5

Now, the slope would be the coefficient of x. Let's denote these m1 and m2.
m1 = -1, m2 = 1.
m1 is not equal to m2, so the lines aren't paralle.
m1 IS the negative reciprocal of m2 ( m1 = (-1)(1/m2)), so the lines are perpendicular.

2. 2y + 3x = 5
3y - 2x = 5

First step: convert to slope-intercept form y = mx + b.

2y + 3x = 5 ---> 2y = -3x + 5 ---> y = (-3/2)x + (5/2)
3y - 2x = 5 ---> 3y = 2x + 5 ---> y = (2/3)x + (5/3)

m1 = -3/2, m2 = 2/3
Negative reciprocal of m1 = (-1)(1/[-3/2]) = 2/3 = m2
Therefore, the lines of question 2 are perpendicular.

3. (1/2)x + (1/3)y = 2
2x + 3y = 4

Slope-intercept form:

L1: (1/2)x + (1/3)y = 2
(1/3)y = (-1/2)x + 2. Multiplying both sides by 3,
y = (-3/2)x + 6

L2: 2x + 3y = 4
3y = -2x + 4
y = (-2/3)x + (4/3)

m1 = -3/2, and the negative reciprocal of m1 is 2/3, which is NOT equal to m2. Therefore, not perpendicular.

m1 is clearly not equal to m2 (-3/2 is not equal to -2/3), so the lines are not parallel.

Therefore, they are neither parallel nor perpendicular

Answer 3

Rule of thumb: In the subject of equations of lines,
parallel = "same slope"
perpendicular = "negative reciprocal"

That's exactly how we solve these; we find the slope m, and compare them. Note that negative reciprocal means the reciprocal of the slope in question, multiplied by -1.

1. x + y = 5, x - y = 5

The first step would be to convert your lines to slope-intercept for; this is of the form y = mx + b.

y = -x + 5, y = x - 5

Now, the slope would be the coefficient of x. Let's denote these m1 and m2.
m1 = -1, m2 = 1.
m1 is not equal to m2, so the lines aren't paralle.
m1 IS the negative reciprocal of m2 ( m1 = (-1)(1/m2)), so the lines are perpendicular.

2. 2y + 3x = 5
3y - 2x = 5

First step: convert to slope-intercept form y = mx + b.

2y + 3x = 5 ---> 2y = -3x + 5 ---> y = (-3/2)x + (5/2)
3y - 2x = 5 ---> 3y = 2x + 5 ---> y = (2/3)x + (5/3)

m1 = -3/2, m2 = 2/3
Negative reciprocal of m1 = (-1)(1/[-3/2]) = 2/3 = m2
Therefore, the lines of question 2 are perpendicular.

3. (1/2)x + (1/3)y = 2
2x + 3y = 4

Slope-intercept form:

L1: (1/2)x + (1/3)y = 2
(1/3)y = (-1/2)x + 2. Multiplying both sides by 3,
y = (-3/2)x + 6

L2: 2x + 3y = 4
3y = -2x + 4
y = (-2/3)x + (4/3)

m1 = -3/2, and the negative reciprocal of m1 is 2/3, which is NOT equal to m2. Therefore, not perpendicular.

m1 is clearly not equal to m2 (-3/2 is not equal to -2/3), so the lines are not parallel.

Therefore, they are neither parallel nor perpendicular.

Answer 4

The first thing you have to do to be able to graph these eqations is solve for y, so that you equation fits the y=mx+b format, with m being the slope (rise/run), and b being your y-intercept.

1. x + y = 5 ----> y=5-x (or y= -x+5)
x-y=5 -----> y=x-5

graph this by plugging in x values (for the x axis) which gives you values for the y (y- axis).
You see that by looking at the y=mx+b equation that the 1st equation has a negative slope (-x = (-1)x so... m=-1) and the second has a pos. slope... (x = (1)x so... m= +1)
This means that one graph is increasing while the other is decreasing, and the slopes are the "same", only neg and pos so the graphs are perpendicular.

2. Again, solve for y.

y=- 3/2x +5/2
y= 2/3x+5/3

again look at the slopes; one is neg, the other pos, but they are not the "same" slopes, so they intersects, but are not perpendicular, nor parallel. (Parallel lines would have the same slope).

3. solve for y
y=-3/2x +6
y= -2/3x -4/3

these graphs intersect, but are not at right angles from one another, so not perpendicular or parallel.

Answer 5

the first is parallel.
second neither.
third neither.
cheers