Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**mathland****Member**- Registered: 2021-03-25
- Posts: 444

Determine whether the lines are parallel,

perpendicular, or neither.

Line 1: y = (1/4) x − 1

Line 2: y = 4x + 7

What are the rules that determine if two given lines are

parallel, perpendicular, or neither?

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,275

I started my diagram with a rectangle 3 by 2. Then I drew a second one 6 by 4 and a third 2 by 3 (ie' the same as the first but turned 90 degrees)

What follows would still be true had I drawn a different start rectangle so these results are generally true.

You can see that the lines AC and HF are parallel. The first has gradient 2/3; the second 4/6 = 2/3.

So to test for parallels, check for equal gradients.

The line IK is at right angles to AC. It's gradient is IL/LK = -3/2 (minus because we have to go down that line)

To test for perpendicular lines check if the gradients multiply together to give -1.

This is because, when you turn the rectangle 90 degrees, the 'up' becomes the 'across' and the 'across' becomes minus the 'up'.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

Offline

**mathland****Member**- Registered: 2021-03-25
- Posts: 444

zetafunc wrote:

Suppose that you have a graph with two parallel lines.

What can you say about their slopes?

What is the slope of a straight line with equation y = mx + c?

1. The slopes are negative reciprocal of each other.

2. The slope of the line y = mx + b is the coefficient of x or m.

Offline

**mathland****Member**- Registered: 2021-03-25
- Posts: 444

Bob wrote:

I started my diagram with a rectangle 3 by 2. Then I drew a second one 6 by 4 and a third 2 by 3 (ie' the same as the first but turned 90 degrees)

What follows would still be true had I drawn a different start rectangle so these results are generally true.

https://i.imgur.com/xe2ra3X.gif

You can see that the lines AC and HF are parallel. The first has gradient 2/3; the second 4/6 = 2/3.

So to test for parallels, check for equal gradients.

The line IK is at right angles to AC. It's gradient is IL/LK = -3/2 (minus because we have to go down that line)

To test for perpendicular lines check if the gradients multiply together to give -1.

This is because, when you turn the rectangle 90 degrees, the 'up' becomes the 'across' and the 'across' becomes minus the 'up'.

Bob

1. The slopes are negative reciprocal of each other.

2. The slope of the line y = mx + b is the coefficient of x or m.

Offline

mathland wrote:

2. The slope of the line y = mx + b is the coefficient of x or m.

Correct -- although it's better to say that it's the coefficient of x, as m doesn't usually represent a variable in the equation of a straight line.

mathland wrote:

1. The slopes are negative reciprocal of each other.

No, that's only true for perpendicular lines.

Bob has helpfully provided a picture to illustrate what is going on -- have you read and understood his explanation?

Offline

**mathland****Member**- Registered: 2021-03-25
- Posts: 444

zetafunc wrote:

mathland wrote:2. The slope of the line y = mx + b is the coefficient of x or m.

Correct -- although it's better to say that it's the coefficient of x, as m doesn't usually represent a variable in the equation of a straight line.

mathland wrote:1. The slopes are negative reciprocal of each other.

No, that's only true for perpendicular lines.

Bob has helpfully provided a picture to illustrate what is going on -- have you read and understood his explanation?

I read Bob's reply twice and carefully studied his geometric interpretation but cannot make sense of the situation. Can you provide further insight?

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,275

Can you view this diagram?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

Offline

**mathland****Member**- Registered: 2021-03-25
- Posts: 444

Bob wrote:

Can you view this diagram?

https://i.imgur.com/xe2ra3X.gif

Bob

I can view the diagram but cannot make sense in terms of finding the answer.

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,275

Ok. I'll try again.

The slope or gradient of a line is found by noting the coordinates of two points and working out [difference in ys]/[difference in xs]

So I choose a line, AC, and then boxed it in with a rectangle ABCD, making the sides horizontal and vertical.

The gradient of AC is CD/AD which in my diagram is 2/3

Then I turned the rectangle around 90 degrees to make IJKL. The diagonal IK is at right angles to AC. Its gradient is IL/LK.

In going from L to K we are going in the positive direction so we can record that distance as +2. But we have to go down to get from I to L, so rather than recording this distance as +3, it should be written as -3. That makes the gradient -3/2. This is what you should expect as it slopes the other way, so it should have a negative gradient.

Note that 2/3 multiplied by -3/2 gives -1.

Of course my diagram only shows this for those measurements but you should be able to see that it will always work out like this whatever the size of the rectangle because the up in one rectangle becomes the across in the other.

So the general test for perpendicular lines is to ask if the product of the gradients is -1.

My third rectangle has HF parallel to AC. I've made the rectangle twice the size but, as both the up and the across have got bigger by the same factor, the gradient comes out the same as AC.

Again this is a general rule: same gradient means parallel.

Hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

Offline

**mathland****Member**- Registered: 2021-03-25
- Posts: 444

Bob wrote:

Ok. I'll try again.

The slope or gradient of a line is found by noting the coordinates of two points and working out [difference in ys]/[difference in xs]

So I choose a line, AC, and then boxed it in with a rectangle ABCD, making the sides horizontal and vertical.

The gradient of AC is CD/AD which in my diagram is 2/3

Then I turned the rectangle around 90 degrees to make IJKL. The diagonal IK is at right angles to AC. Its gradient is IL/LK.

In going from L to K we are going in the positive direction so we can record that distance as +2. But we have to go down to get from I to L, so rather than recording this distance as +3, it should be written as -3. That makes the gradient -3/2. This is what you should expect as it slopes the other way, so it should have a negative gradient.

Note that 2/3 multiplied by -3/2 gives -1.

Of course my diagram only shows this for those measurements but you should be able to see that it will always work out like this whatever the size of the rectangle because the up in one rectangle becomes the across in the other.

So the general test for perpendicular lines is to ask if the product of the gradients is -1.

My third rectangle has HF parallel to AC. I've made the rectangle twice the size but, as both the up and the across have got bigger by the same factor, the gradient comes out the same as AC.

Again this is a general rule: same gradient means parallel.

Hope that helps,

Bob

Much better. Thanks.

Offline

Pages: **1**