Math / Algebra

Question4. Classify the following pairs of equations given as parallel, perpendicular, or neither on the basis of their respected slopes. a) $\begin{array}{l} y=\frac{2}{3} x-8 \\ y=\frac{7}{4} x+8 \end{array}$ b) $\begin{array}{l} y=\frac{7}{4} x+9 \\ y=-\frac{4}{7} x-1 \end{array}$ c) $-2 x-5 y=20$ $10 x+9 y=-9$

Studdy Solution

STEP 1

What is this asking? We need to figure out if some lines are parallel, perpendicular, or neither by looking at how steep they are. Watch out! Don't mix up parallel and perpendicular!
Parallel lines have the same slope, like train tracks.
Perpendicular lines meet at a right angle, and their slopes are negative reciprocals of each other.

STEP 2

1. Analyze the first pair of equations.
2. Analyze the second pair of equations.
3. Analyze the third pair of equations.

STEP 3

Alright, let's look at the first pair!
We've got $y = \frac{2}{3}x - 8$ and $y = \frac{7}{4}x + 8$ .
The slope of the first line is $\frac{2}{3}$ , and the slope of the second line is $\frac{7}{4}$ .

STEP 4

Are they the same?
Nope! Are they negative reciprocals?
Meaning, if we flip one fraction and multiply by $-1$ , do we get the other?
Let's check: $-1 \cdot \frac{3}{2} = -\frac{3}{2}$ .
This isn't $\frac{7}{4}$ , so they're not perpendicular either.

STEP 5

So, these lines are neither parallel nor perpendicular!

STEP 6

Next up: $y = \frac{7}{4}x + 9$ and $y = -\frac{4}{7}x - 1$ .
The slope of the first line is $\frac{7}{4}$ , and the slope of the second line is $-\frac{4}{7}$ .

STEP 7

They're not the same, so they're not parallel.
But are they negative reciprocals?
Let's see: $-1 \cdot \frac{4}{7} = -\frac{4}{7}$ .
Hey, that matches the slope of the second line!

STEP 8

These lines are perpendicular!
They meet at a perfect right angle.

STEP 9

Last ones!
We have $-2x - 5y = 20$ and $10x + 9y = -9$ .
These aren't in slope-intercept form, so we need to rewrite them.

STEP 10

Let's start with $-2x - 5y = 20$ .
We want to isolate $y$ . Add $2x$ to both sides to get $-5y = 2x + 20$ .
Now, divide both sides by $-5$ to get $y = -\frac{2}{5}x - 4$ .
The slope is $-\frac{2}{5}$ .

STEP 11

Now for $10x + 9y = -9$ . Subtract $10x$ from both sides to get $9y = -10x - 9$ . Divide both sides by $9$ to get $y = -\frac{10}{9}x - 1$ .
The slope is $-\frac{10}{9}$ .

STEP 12

The slopes aren't the same, so they're not parallel.
Let's check for perpendicularity. $-1 \cdot \frac{5}{2} = \frac{5}{2}$ .
This isn't $-\frac{10}{9}$ , so they're not perpendicular either.

Studdy Solution

STEP 1

STEP 2

1. Analyze the first pair of equations.
2. Analyze the second pair of equations.
3. Analyze the third pair of equations.

STEP 3

Alright, let's look at the first pair!
We've got $y = \frac{2}{3}x - 8$ and $y = \frac{7}{4}x + 8$ .
The slope of the first line is $\frac{2}{3}$ , and the slope of the second line is $\frac{7}{4}$ .

STEP 4

Are they the same?
Nope! Are they negative reciprocals?
Meaning, if we flip one fraction and multiply by $-1$ , do we get the other?
Let's check: $-1 \cdot \frac{3}{2} = -\frac{3}{2}$ .
This isn't $\frac{7}{4}$ , so they're not perpendicular either.

STEP 5

So, these lines are neither parallel nor perpendicular!

STEP 6

Next up: $y = \frac{7}{4}x + 9$ and $y = -\frac{4}{7}x - 1$ .
The slope of the first line is $\frac{7}{4}$ , and the slope of the second line is $-\frac{4}{7}$ .

STEP 7

They're not the same, so they're not parallel.
But are they negative reciprocals?
Let's see: $-1 \cdot \frac{4}{7} = -\frac{4}{7}$ .
Hey, that matches the slope of the second line!

STEP 8

These lines are perpendicular!
They meet at a perfect right angle.

STEP 9

Last ones!
We have $-2x - 5y = 20$ and $10x + 9y = -9$ .
These aren't in slope-intercept form, so we need to rewrite them.

STEP 10

Let's start with $-2x - 5y = 20$ .
We want to isolate $y$ . Add $2x$ to both sides to get $-5y = 2x + 20$ .
Now, divide both sides by $-5$ to get $y = -\frac{2}{5}x - 4$ .
The slope is $-\frac{2}{5}$ .

STEP 11

Now for $10x + 9y = -9$ . Subtract $10x$ from both sides to get $9y = -10x - 9$ . Divide both sides by $9$ to get $y = -\frac{10}{9}x - 1$ .
The slope is $-\frac{10}{9}$ .

STEP 12

The slopes aren't the same, so they're not parallel.
Let's check for perpendicularity. $-1 \cdot \frac{5}{2} = \frac{5}{2}$ .
This isn't $-\frac{10}{9}$ , so they're not perpendicular either.

STEP 13

These lines are neither parallel nor perpendicular!

STEP 14

a) Neither b) Perpendicular c) Neither

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Studdy Solution

STEP 1

STEP 2

1. Analyze the first pair of equations.2. Analyze the second pair of equations.3. Analyze the third pair of equations.

STEP 3

Alright, let's **look** at the first pair!We've got y=23x−8y = \frac{2}{3}x - 8y=32​x−8 and y=74x+8y = \frac{7}{4}x + 8y=47​x+8.The **slope** of the first line is 23\frac{2}{3}32​, and the **slope** of the second line is 74\frac{7}{4}47​.

STEP 4

Are they the same?Nope! Are they negative reciprocals?Meaning, if we flip one fraction and multiply by −1-1−1, do we get the other?Let's **check**: −1⋅32=−32-1 \cdot \frac{3}{2} = -\frac{3}{2}−1⋅23​=−23​.This isn't 74\frac{7}{4}47​, so they're not perpendicular either.

STEP 5

So, these lines are **neither** parallel nor perpendicular!

STEP 6

Next up: y=74x+9y = \frac{7}{4}x + 9y=47​x+9 and y=−47x−1y = -\frac{4}{7}x - 1y=−74​x−1.The **slope** of the first line is 74\frac{7}{4}47​, and the **slope** of the second line is −47-\frac{4}{7}−74​.

STEP 7

They're not the same, so they're not parallel.But are they negative reciprocals?Let's **see**: −1⋅47=−47-1 \cdot \frac{4}{7} = -\frac{4}{7}−1⋅74​=−74​.Hey, that matches the slope of the second line!

STEP 8

These lines are **perpendicular**!They meet at a perfect right angle.

STEP 9

Last ones!We have −2x−5y=20-2x - 5y = 20−2x−5y=20 and 10x+9y=−910x + 9y = -910x+9y=−9.These aren't in slope-intercept form, so we need to **rewrite** them.

STEP 10

STEP 11

Now for 10x+9y=−910x + 9y = -910x+9y=−9. **Subtract** 10x10x10x from both sides to get 9y=−10x−99y = -10x - 99y=−10x−9. **Divide** both sides by 999 to get y=−109x−1y = -\frac{10}{9}x - 1y=−910​x−1.The **slope** is −109-\frac{10}{9}−910​.

STEP 12

The slopes aren't the same, so they're not parallel.Let's **check** for perpendicularity. −1⋅52=52-1 \cdot \frac{5}{2} = \frac{5}{2}−1⋅25​=25​.This isn't −109-\frac{10}{9}−910​, so they're not perpendicular either.

STEP 13

These lines are **neither** parallel nor perpendicular!

STEP 14

a) **Neither** b) **Perpendicular** c) **Neither**

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1. Analyze the first pair of equations.
2. Analyze the second pair of equations.
3. Analyze the third pair of equations.

Alright, let's look at the first pair!
We've got $y = \frac{2}{3}x - 8$ and $y = \frac{7}{4}x + 8$ .
The slope of the first line is $\frac{2}{3}$ , and the slope of the second line is $\frac{7}{4}$ .

Are they the same?
Nope! Are they negative reciprocals?
Meaning, if we flip one fraction and multiply by $-1$ , do we get the other?
Let's check: $-1 \cdot \frac{3}{2} = -\frac{3}{2}$ .
This isn't $\frac{7}{4}$ , so they're not perpendicular either.

So, these lines are neither parallel nor perpendicular!

Next up: $y = \frac{7}{4}x + 9$ and $y = -\frac{4}{7}x - 1$ .
The slope of the first line is $\frac{7}{4}$ , and the slope of the second line is $-\frac{4}{7}$ .

They're not the same, so they're not parallel.
But are they negative reciprocals?
Let's see: $-1 \cdot \frac{4}{7} = -\frac{4}{7}$ .
Hey, that matches the slope of the second line!

These lines are perpendicular!
They meet at a perfect right angle.

Last ones!
We have $-2x - 5y = 20$ and $10x + 9y = -9$ .
These aren't in slope-intercept form, so we need to rewrite them.

Now for $10x + 9y = -9$ . Subtract $10x$ from both sides to get $9y = -10x - 9$ . Divide both sides by $9$ to get $y = -\frac{10}{9}x - 1$ .
The slope is $-\frac{10}{9}$ .

The slopes aren't the same, so they're not parallel.
Let's check for perpendicularity. $-1 \cdot \frac{5}{2} = \frac{5}{2}$ .
This isn't $-\frac{10}{9}$ , so they're not perpendicular either.

These lines are neither parallel nor perpendicular!

a) Neither b) Perpendicular c) Neither