Math / Algebra

QuestionDetermine whether the graphs of the given equetions are permilet, perpendicular, or metilher. Repilain. 13. $\begin{array}{l} y=x+11 \\ y=-x+2 \end{array}$
16. $y-4=3(x+2)$ $2 x+6 y=10$
14. $\begin{aligned} y & =\frac{3}{4} x-1 \\ y & =\frac{3}{4} x+29\end{aligned}$
17. $y=-7$ $x=2$
15. $y=-2 x+3$ $2 x+y=7$
18. $y=4 x-2$ $-x+4 y=0$
Write an equation in slope-intercept form of the line that passes throught the given point and is perpendicular to the graph of the given equation.

Studdy Solution

STEP 1

What is this asking? We're checking if lines are parallel, perpendicular, or neither, and then writing equations for lines perpendicular to a given line through a specific point. Watch out! Don't mix up parallel and perpendicular!
Parallel lines have the same slope, while perpendicular lines have negative reciprocal slopes.

STEP 2

1. Analyze the first set of equations.
2. Analyze the second set of equations.
3. Tackle the perpendicular line equation problem.

STEP 3

Alright, let's look at problem 13!
We've got $y = x + 11$ and $y = -x + 2$ .
Remember, the slope is the number multiplying $x$ .

STEP 4

In the first equation, the slope is 1.
In the second, it's -1.
Since 1 and -1 are negative reciprocals of each other (meaning $1 \cdot -1 = -1$ ), these lines are perpendicular!

STEP 5

Next up, problem 14: $y = \frac{3}{4}x - 1$ and $y = \frac{3}{4}x + 29$ .

STEP 6

Both equations have a slope of $\frac{\mathbf{3}}{\mathbf{4}}$ .
Same slope?
That means these lines are parallel!

STEP 7

Now for problem 15: $y = -2x + 3$ and $2x + y = 7$ .
We need to rewrite the second equation in slope-intercept form.

STEP 8

Subtracting $2x$ from both sides of $2x + y = 7$ gives us $y = -2x + 7$ .

STEP 9

Now we can see both lines have a slope of $\mathbf{-2}$ .
So, these lines are also parallel!

STEP 10

Problem 16: $y - 4 = 3(x + 2)$ and $2x + 6y = 10$ .
Let's get both into slope-intercept form.

STEP 11

First equation: distribute the 3 to get $y - 4 = 3x + 6$ , then add 4 to both sides: $y = 3x + 10$ .
The slope is 3.
Second equation: subtract $2x$ from both sides to get $6y = -2x + 10$ , then divide by 6: $y = -\frac{1}{3}x + \frac{5}{3}$ .
The slope is $-\frac{\mathbf{1}}{\mathbf{3}}$ .

STEP 12

Since 3 and $-\frac{1}{3}$ are negative reciprocals, these lines are perpendicular!

STEP 13

Problem 17: $y = -7$ and $x = 2$ . $y = -7$ is a horizontal line with a slope of 0, and $x = 2$ is a vertical line with an undefined slope.

STEP 14

Horizontal and vertical lines are always perpendicular!

STEP 15

Lastly, problem 18: $y = 4x - 2$ and $-x + 4y = 0$ .
Let's rewrite the second equation.

STEP 16

Add $x$ to both sides of $-x + 4y = 0$ to get $4y = x$ , then divide by 4: $y = \frac{1}{4}x$ .

STEP 17

The first line has a slope of 4, the second has a slope of $\frac{\mathbf{1}}{\mathbf{4}}$ .
Since these are not negative reciprocals, these lines are neither parallel nor perpendicular.

STEP 18

Problem 13: Perpendicular.
Problem 14: Parallel.
Problem 15: Parallel.
Problem 16: Perpendicular.
Problem 17: Perpendicular.
Problem 18: Neither.

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Studdy Solution

STEP 1

STEP 2

1. Analyze the first set of equations.
2. Analyze the second set of equations.
3. Tackle the perpendicular line equation problem.

STEP 3

Alright, let's look at problem 13!
We've got $y = x + 11$ and $y = -x + 2$ .
Remember, the slope is the number multiplying $x$ .

STEP 4

In the first equation, the slope is 1.
In the second, it's -1.
Since 1 and -1 are negative reciprocals of each other (meaning $1 \cdot -1 = -1$ ), these lines are perpendicular!

STEP 5

Next up, problem 14: $y = \frac{3}{4}x - 1$ and $y = \frac{3}{4}x + 29$ .

STEP 6

Both equations have a slope of $\frac{\mathbf{3}}{\mathbf{4}}$ .
Same slope?
That means these lines are parallel!

STEP 7

Now for problem 15: $y = -2x + 3$ and $2x + y = 7$ .
We need to rewrite the second equation in slope-intercept form.

STEP 8

Subtracting $2x$ from both sides of $2x + y = 7$ gives us $y = -2x + 7$ .

STEP 9

Now we can see both lines have a slope of $\mathbf{-2}$ .
So, these lines are also parallel!

STEP 10

Problem 16: $y - 4 = 3(x + 2)$ and $2x + 6y = 10$ .
Let's get both into slope-intercept form.

STEP 11

First equation: distribute the 3 to get $y - 4 = 3x + 6$ , then add 4 to both sides: $y = 3x + 10$ .
The slope is 3.
Second equation: subtract $2x$ from both sides to get $6y = -2x + 10$ , then divide by 6: $y = -\frac{1}{3}x + \frac{5}{3}$ .
The slope is $-\frac{\mathbf{1}}{\mathbf{3}}$ .

STEP 12

Since 3 and $-\frac{1}{3}$ are negative reciprocals, these lines are perpendicular!

STEP 13

Problem 17: $y = -7$ and $x = 2$ . $y = -7$ is a horizontal line with a slope of 0, and $x = 2$ is a vertical line with an undefined slope.

STEP 14

Horizontal and vertical lines are always perpendicular!

STEP 15

Lastly, problem 18: $y = 4x - 2$ and $-x + 4y = 0$ .
Let's rewrite the second equation.

STEP 16

Add $x$ to both sides of $-x + 4y = 0$ to get $4y = x$ , then divide by 4: $y = \frac{1}{4}x$ .

STEP 17

The first line has a slope of 4, the second has a slope of $\frac{\mathbf{1}}{\mathbf{4}}$ .
Since these are not negative reciprocals, these lines are neither parallel nor perpendicular.

STEP 18

Problem 13: Perpendicular.
Problem 14: Parallel.
Problem 15: Parallel.
Problem 16: Perpendicular.
Problem 17: Perpendicular.
Problem 18: Neither.

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Studdy Solution

STEP 1

STEP 2

1. Analyze the first set of equations.2. Analyze the second set of equations.3. Tackle the perpendicular line equation problem.

STEP 3

Alright, let's look at problem 13!We've got y=x+11y = x + 11y=x+11 and y=−x+2y = -x + 2y=−x+2.Remember, the **slope** is the number multiplying xxx.

STEP 4

In the first equation, the slope is **1**.In the second, it's **-1**.Since 1 and -1 are negative reciprocals of each other (meaning 1⋅−1=−11 \cdot -1 = -11⋅−1=−1), these lines are **perpendicular**!

STEP 5

Next up, problem 14: y=34x−1y = \frac{3}{4}x - 1y=43​x−1 and y=34x+29y = \frac{3}{4}x + 29y=43​x+29.

STEP 6

Both equations have a slope of 34\frac{\mathbf{3}}{\mathbf{4}}43​.Same slope?That means these lines are **parallel**!

STEP 7

Now for problem 15: y=−2x+3y = -2x + 3y=−2x+3 and 2x+y=72x + y = 72x+y=7.We need to rewrite the second equation in slope-intercept form.

STEP 8

Subtracting 2x2x2x from both sides of 2x+y=72x + y = 72x+y=7 gives us y=−2x+7y = -2x + 7y=−2x+7.

STEP 9

Now we can see both lines have a slope of −2\mathbf{-2}−2.So, these lines are also **parallel**!

STEP 10

Problem 16: y−4=3(x+2)y - 4 = 3(x + 2)y−4=3(x+2) and 2x+6y=102x + 6y = 102x+6y=10.Let's get both into slope-intercept form.

STEP 11

STEP 12

Since 3 and −13-\frac{1}{3}−31​ are negative reciprocals, these lines are **perpendicular**!

STEP 13

Problem 17: y=−7y = -7y=−7 and x=2x = 2x=2. y=−7y = -7y=−7 is a horizontal line with a slope of **0**, and x=2x = 2x=2 is a vertical line with an *undefined* slope.

STEP 14

Horizontal and vertical lines are always **perpendicular**!

STEP 15

Lastly, problem 18: y=4x−2y = 4x - 2y=4x−2 and −x+4y=0-x + 4y = 0−x+4y=0.Let's rewrite the second equation.

STEP 16

Add xxx to both sides of −x+4y=0-x + 4y = 0−x+4y=0 to get 4y=x4y = x4y=x, then divide by 4: y=14xy = \frac{1}{4}xy=41​x.

STEP 17

The first line has a slope of **4**, the second has a slope of 14\frac{\mathbf{1}}{\mathbf{4}}41​.Since these are *not* negative reciprocals, these lines are **neither** parallel nor perpendicular.

STEP 18

Problem 13: **Perpendicular**.Problem 14: **Parallel**.Problem 15: **Parallel**.Problem 16: **Perpendicular**.Problem 17: **Perpendicular**.Problem 18: **Neither**.

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1. Analyze the first set of equations.
2. Analyze the second set of equations.
3. Tackle the perpendicular line equation problem.

Alright, let's look at problem 13!
We've got $y = x + 11$ and $y = -x + 2$ .
Remember, the slope is the number multiplying $x$ .

In the first equation, the slope is 1.
In the second, it's -1.
Since 1 and -1 are negative reciprocals of each other (meaning $1 \cdot -1 = -1$ ), these lines are perpendicular!

Next up, problem 14: $y = \frac{3}{4}x - 1$ and $y = \frac{3}{4}x + 29$ .

Both equations have a slope of $\frac{\mathbf{3}}{\mathbf{4}}$ .
Same slope?
That means these lines are parallel!

Now for problem 15: $y = -2x + 3$ and $2x + y = 7$ .
We need to rewrite the second equation in slope-intercept form.

Subtracting $2x$ from both sides of $2x + y = 7$ gives us $y = -2x + 7$ .

Now we can see both lines have a slope of $\mathbf{-2}$ .
So, these lines are also parallel!

Problem 16: $y - 4 = 3(x + 2)$ and $2x + 6y = 10$ .
Let's get both into slope-intercept form.

Since 3 and $-\frac{1}{3}$ are negative reciprocals, these lines are perpendicular!

Problem 17: $y = -7$ and $x = 2$ . $y = -7$ is a horizontal line with a slope of 0, and $x = 2$ is a vertical line with an undefined slope.

Horizontal and vertical lines are always perpendicular!

Lastly, problem 18: $y = 4x - 2$ and $-x + 4y = 0$ .
Let's rewrite the second equation.

Add $x$ to both sides of $-x + 4y = 0$ to get $4y = x$ , then divide by 4: $y = \frac{1}{4}x$ .

The first line has a slope of 4, the second has a slope of $\frac{\mathbf{1}}{\mathbf{4}}$ .
Since these are not negative reciprocals, these lines are neither parallel nor perpendicular.

Problem 13: Perpendicular.
Problem 14: Parallel.
Problem 15: Parallel.
Problem 16: Perpendicular.
Problem 17: Perpendicular.
Problem 18: Neither.