find the slope of a line perpendicular to each given line.
11) ( y = -5x + 2 )
12) ( y = \frac{1}{2}x + 1 )

find the slope of a line parallel to each given line.
13) ( y = -x + 5 )
14) ( x = 3 )

find the slope of the line through each pair of points.
15) ( (14, 13), (4, -1) )
16) ( (13, 14), (10, 2) )

17) ( mangle qrc = 6x + 2 ), ( mangle crs = 85^circ ), and ( mangle qrs = 21x - 3 ). find ( x ).
18) ( mangle pfe = x + 58 ), ( mangle gfp = x + 103 ), and ( mangle gfe = 155^circ ). find ( mangle gfp ).

find the measure of the angle indicated in bold.
19) (diagram of two parallel lines cut by a transversal, angles ( 6x + 7 ) and ( 7x - 1 ))
20) (diagram of two horizontal lines cut by a vertical transversal, angles ( 41x + 3 ) and ( 43x - 1 ))

Question

find the slope of a line perpendicular to each given line.
11) ( y = -5x + 2 )
12) ( y = \frac{1}{2}x + 1 )

find the slope of a line parallel to each given line.
13) ( y = -x + 5 )
14) ( x = 3 )

find the slope of the line through each pair of points.
15) ( (14, 13), (4, -1) )
16) ( (13, 14), (10, 2) )

17) ( mangle qrc = 6x + 2 ), ( mangle crs = 85^circ ), and ( mangle qrs = 21x - 3 ). find ( x ).
18) ( mangle pfe = x + 58 ), ( mangle gfp = x + 103 ), and ( mangle gfe = 155^circ ). find ( mangle gfp ).

find the measure of the angle indicated in bold.
19) (diagram of two parallel lines cut by a transversal, angles ( 6x + 7 ) and ( 7x - 1 ))
20) (diagram of two horizontal lines cut by a vertical transversal, angles ( 41x + 3 ) and ( 43x - 1 ))

Accepted Answer

### Problem 11: Find the slope of a line perpendicular to $ y = -5x + 2 $
# Explanation:
## Step 1: Identify the slope of the given line
The given line is in slope - intercept form $ y=mx + b $, where $ m $ is the slope. For the line $ y=-5x + 2 $, the slope of the given line $ m_1=-5 $.

## Step 2: Find the slope of the perpendicular line
If two lines are perpendicular, the product of their slopes is $ - 1 $, i.e., $ m_1\times m_2=-1 $. Let the slope of the perpendicular line be $ m_2 $. We have $ -5\times m_2=-1 $. Solving for $ m_2 $, we get $ m_2=\frac{-1}{-5}=\frac{1}{5} $.
# Answer:
$\frac{1}{5}$

### Problem 12: Find the slope of a line perpendicular to $ y=\frac{1}{2}x + 1 $
# Explanation:
## Step 1: Identify the slope of the given line
For the line $ y = \frac{1}{2}x+1 $ (in slope - intercept form $ y = mx + b $), the slope of the given line $ m_1=\frac{1}{2} $.

## Step 2: Find the slope of the perpendicular line
Using the property of perpendicular lines $ m_1\times m_2=-1 $. Let $ m_2 $ be the slope of the perpendicular line. Then $ \frac{1}{2}\times m_2=-1 $. Solving for $ m_2 $, we multiply both sides by 2: $ m_2=-2 $.
# Answer:
$-2$

### Problem 13: Find the slope of a line parallel to $ y=-x + 5 $
# Explanation:
## Step 1: Identify the slope of the given line
The line $ y=-x + 5 $ is in slope - intercept form $ y=mx + b $, where $ m=-1 $.

## Step 2: Find the slope of the parallel line
If two lines are parallel, they have the same slope. So the slope of the line parallel to $ y=-x + 5 $ is also $ - 1 $.
# Answer:
$-1$

### Problem 14: Find the slope of a line parallel to $ x = 3 $
# Explanation:
## Step 1: Analyze the line $ x = 3 $
The line $ x = 3 $ is a vertical line. Vertical lines have an undefined slope.

## Step 2: Determine the slope of the parallel line
A line parallel to a vertical line is also a vertical line, so it also has an undefined slope.
# Answer:
Undefined

### Problem 15: Find the slope of the line through $ (14,13) $ and $ (4,-1) $
# Explanation:
## Step 1: Recall the slope formula
The slope $ m $ between two points $ (x_1,y_1) $ and $ (x_2,y_2) $ is given by $ m=\frac{y_2 - y_1}{x_2 - x_1} $. Let $ (x_1,y_1)=(14,13) $ and $ (x_2,y_2)=(4,-1) $.

## Step 2: Substitute the values into the formula
$ m=\frac{-1 - 13}{4 - 14}=\frac{-14}{-10}=\frac{7}{5} $
# Answer:
$\frac{7}{5}$

### Problem 16: Find the slope of the line through $ (13,14) $ and $ (10,2) $
# Explanation:
## Step 1: Recall the slope formula
The slope $ m $ between two points $ (x_1,y_1) $ and $ (x_2,y_2) $ is $ m=\frac{y_2 - y_1}{x_2 - x_1} $. Let $ (x_1,y_1)=(13,14) $ and $ (x_2,y_2)=(10,2) $.

## Step 2: Substitute the values into the formula
$ m=\frac{2 - 14}{10 - 13}=\frac{-12}{-3}=4 $
# Answer:
$4$

### Problem 17: Solve for $ x $ given $ m\angle QRC = 6x + 2 $, $ m\angle CRS=85^{\circ} $, and $ m\angle QRS = 21x-3 $
# Explanation:
## Step 1: Analyze the angle relationship
From the diagram, $ \angle QRS=\angle QRC+\angle CRS $. So $ m\angle QRS=m\angle QRC + m\angle CRS $.

## Step 2: Substitute the given angle measures
Substitute $ m\angle QRC = 6x + 2 $, $ m\angle CRS = 85^{\circ} $, and $ m\angle QRS=21x - 3 $ into the equation: $ 21x-3=(6x + 2)+85 $.

## Step 3: Simplify and solve for $ x $
First, simplify the right - hand side: $ 21x-3=6x+87 $. Subtract $ 6x $ from both sides: $ 21x-6x-3=6x - 6x+87 $, which gives $ 15x-3 = 87 $. Then add 3 to both sides: $ 15x-3 + 3=87 + 3 $, so $ 15x=90 $. Divide both sides by 15: $ x=\frac{90}{15}=6 $.
# Answer:
$6$

### Problem 18: Find $ m\angle GFP $ given $ m\angle PFE=x + 58 $, $ m\angle GFP=x + 103 $, and $ m\angle GFE = 155^{\circ} $
# Explanation:
## Step 1: Analyze the angle relationship
From the diagram, $ \angle GFE=\angle GFP+\angle PFE $. So $ m\angle GFE=m\angle GFP + m\angle PFE $.

## Step 2: Substitute the given angle measures
Substitute $ m\angle PFE=x + 58 $, $ m\angle GFP=x + 103 $, and $ m\angle GFE = 155^{\circ} $ into the equation: $ 155=(x + 103)+(x + 58) $.

## Step 3: Simplify and solve for $ x $
Simplify the right - hand side: $ 155=2x+161 $. Subtract 161 from both sides: $ 155 - 161=2x+161 - 161 $, which gives $ - 6 = 2x $. Divide both sides by 2: $ x=-3 $.

## Step 4: Find $ m\angle GFP $
Substitute $ x=-3 $ into $ m\angle GFP=x + 103 $: $ m\angle GFP=-3 + 103 = 100^{\circ} $.
# Answer:
$100^{\circ}$

### Problem 19: Find the measure of the angle indicated (the angles $ 6x + 7 $ and $ 7x-1 $ are corresponding angles, so they are equal)
# Explanation:
## Step 1: Set up the equation
Since the two lines are parallel and cut by a transversal, the corresponding angles are equal. So $ 6x + 7=7x-1 $.

## Step 2: Solve for $ x $
Subtract $ 6x $ from both sides: $ 6x+7-6x=7x - 1-6x $, which gives $ 7=x - 1 $. Add 1 to both sides: $ x=8 $.

## Step 3: Find the measure of the angle
Substitute $ x = 8 $ into $ 6x+7 $: $ 6\times8+7=48 + 7=55^{\circ} $ (we could also substitute into $ 7x - 1 $: $ 7\times8-1=56 - 1 = 55^{\circ} $).
# Answer:
$55^{\circ}$

### Problem 20: Find the measure of the angle indicated (the two angles $ 41x + 3 $ and $ 43x-1 $ are vertical angles? No, they are adjacent and form a linear pair? Wait, no, since the lines are parallel and cut by a vertical transversal, the angles $ 41x + 3 $ and $ 43x-1 $ are equal? Wait, no, actually, since the transversal is vertical, the two angles $ 41x + 3 $ and $ 43x-1 $ are equal (alternate interior angles or corresponding angles). So we set $ 41x+3=43x - 1 $.

# Explanation:
## Step 1: Set up the equation
$ 41x+3=43x - 1 $

## Step 2: Solve for $ x $
Subtract $ 41x $ from both sides: $ 3 = 2x-1 $. Add 1 to both sides: $ 4 = 2x $. Divide both sides by 2: $ x = 2 $.

## Step 3: Find the measure of the angle
Substitute $ x = 2 $ into $ 41x+3 $: $ 41\times2+3=82 + 3=85^{\circ} $ (we could also substitute into $ 43x-1 $: $ 43\times2-1=86 - 1 = 85^{\circ} $).
# Answer:
$85^{\circ}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 11: Find the slope of a line perpendicular to \( y = -5x + 2 \)

Step 1: Identify the slope of the given line

Step 2: Find the slope of the perpendicular line

Step 1: Identify the slope of the given line

Step 2: Find the slope of the perpendicular line

Step 1: Identify the slope of the given line

Step 2: Find the slope of the parallel line

Answer:

Problem 12: Find the slope of a line perpendicular to \( y=\frac{1}{2}x + 1 \)