im1 parallel & perpendicular lines project
record of evidence c
project
score
name ashley guido
date dec 27 25 period 3
complete the record of evidence. graph the items on the coordinate plane. then color the sections.
1. a. graph the line
y = 2.
b. list two points that are on the line.
( 2 , 0 ) and ( ___ , ___ )
c. is it horizontal or vertical?
2. a. write an equation for the line through the
point (6, −5) that is parallel to y = 2.
b. graph the line.
3. a. write an equation for the line through the point
(6, −3) that is perpendicular to y = 2.
b. graph the line.
4. a. write an equation for a line parallel to the
given line that passes through the point.
y = −3x + 5 ; (1, 7)
b. graph the line.
5. a. write an equation for a line perpendicular to
the given line that passes through the point.
y = $\frac{1}{7}$x + 4 ; (−4, 3)
b. graph the line.
6. a. write an equation for a line parallel to the
given line that passes through the point.
y − 2 = −$\frac{7}{3}$(x + 3) ; (3, −5)
b. graph the line.
7. a. write an equation for a line perpendicular to
the given line that passes through the point.
y + 9 = −$\frac{7}{4}$(x − 6) ; (7, 8)
b. graph the line.
8. a. write an equation for a line parallel to the
given line that passes through the point.
9x + 3y = 6 ; (7, 4)
b. graph the line.
9. a. write an equation for a line perpendicular to
the given line that passes through the point.
10x + 3y = −15 ; (0, 8)
b. graph the line.

Question

im1 parallel & perpendicular lines project
record of evidence c
project
score
name ashley guido
date dec 27 25 period 3
complete the record of evidence. graph the items on the coordinate plane. then color the sections.
1. a. graph the line
   y = 2.
   b. list two points that are on the line.
      ( 2 , 0 ) and ( ___ , ___ )
   c. is it horizontal or vertical?
2. a. write an equation for the line through the
   point (6, −5) that is parallel to y = 2.
   b. graph the line.
3. a. write an equation for the line through the point
   (6, −3) that is perpendicular to y = 2.
   b. graph the line.
4. a. write an equation for a line parallel to the
   given line that passes through the point.
   y = −3x + 5 ; (1, 7)
   b. graph the line.
5. a. write an equation for a line perpendicular to
   the given line that passes through the point.
   y = $\frac{1}{7}$x + 4 ; (−4, 3)
   b. graph the line.
6. a. write an equation for a line parallel to the
   given line that passes through the point.
   y − 2 = −$\frac{7}{3}$(x + 3) ; (3, −5)
   b. graph the line.
7. a. write an equation for a line perpendicular to
   the given line that passes through the point.
   y + 9 = −$\frac{7}{4}$(x − 6) ; (7, 8)
   b. graph the line.
8. a. write an equation for a line parallel to the
   given line that passes through the point.
   9x + 3y = 6 ; (7, 4)
   b. graph the line.
9. a. write an equation for a line perpendicular to
   the given line that passes through the point.
   10x + 3y = −15 ; (0, 8)
   b. graph the line.

Accepted Answer

### Problem 1:
# Explanation:
## Step 1: Understand the line $ y = 2 $
The line $ y = 2 $ is a horizontal line (since it has the form $ y = k $, where $ k $ is a constant) that passes through all points with a $ y $-coordinate of 2.
## Step 2: Find points on the line
Any point with $ y = 2 $ will lie on this line. For example, when $ x = 0 $, the point is $ (0, 2) $; when $ x = 3 $, the point is $ (3, 2) $.
## Step 3: Determine if it's horizontal or vertical
Since the equation is $ y = 2 $ (constant $ y $-value), it is a horizontal line.

# Answer:
b. Two points: $ (0, 2) $ and $ (3, 2) $ (answers may vary, any points with $ y = 2 $ are correct)
c. Horizontal

### Problem 2:
# Explanation:
## Step 1: Recall the slope of parallel lines
Parallel lines have the same slope. The line $ y = 2 $ is horizontal, so its slope $ m = 0 $.
## Step 2: Use the point-slope form
The point-slope form of a line is $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (6, -5) $ and $ m = 0 $. Substituting these values, we get $ y - (-5) = 0(x - 6) $, which simplifies to $ y + 5 = 0 $, or $ y = -5 $.

# Answer:
a. The equation is $ y = -5 $

### Problem 3:
# Explanation:
## Step 1: Recall the slope of perpendicular lines
A line perpendicular to a horizontal line ($ y = 2 $, slope $ 0 $) is a vertical line. The slope of a vertical line is undefined.
## Step 2: Determine the equation of the vertical line
A vertical line passing through $ (6, -3) $ has the equation $ x = 6 $ (since all points on a vertical line have the same $ x $-coordinate).

# Answer:
a. The equation is $ x = 6 $

### Problem 4:
# Explanation:
## Step 1: Identify the slope of the given line
The given line is $ y = -3x + 5 $, which is in slope-intercept form ($ y = mx + b $), so its slope $ m = -3 $.
## Step 2: Recall the slope of parallel lines
Parallel lines have the same slope, so the slope of the new line is also $ -3 $.
## Step 3: Use the point-slope form
Using the point $ (1, 7) $ and $ m = -3 $ in the point-slope form $ y - y_1 = m(x - x_1) $:
$ y - 7 = -3(x - 1) $
Simplify: $ y - 7 = -3x + 3 $, then $ y = -3x + 10 $.

# Answer:
a. The equation is $ y = -3x + 10 $

### Problem 5:
# Explanation:
## Step 1: Identify the slope of the given line
The given line is $ y = \frac{1}{7}x + 4 $, so its slope $ m_1 = \frac{1}{7} $.
## Step 2: Recall the slope of perpendicular lines
The slope of a line perpendicular to a line with slope $ m_1 $ is $ m_2 = -\frac{1}{m_1} $. So $ m_2 = -7 $.
## Step 3: Use the point-slope form
Using the point $ (-4, 3) $ and $ m = -7 $ in the point-slope form $ y - y_1 = m(x - x_1) $:
$ y - 3 = -7(x - (-4)) $
Simplify: $ y - 3 = -7(x + 4) $
$ y - 3 = -7x - 28 $
$ y = -7x - 25 $

# Answer:
a. The equation is $ y = -7x - 25 $

### Problem 6:
# Explanation:
## Step 1: Identify the slope of the given line
The given line is $ y - 2 = -\frac{7}{3}(x + 3) $, which is in point-slope form ($ y - y_1 = m(x - x_1) $), so its slope $ m = -\frac{7}{3} $.
## Step 2: Recall the slope of parallel lines
Parallel lines have the same slope, so the slope of the new line is also $ -\frac{7}{3} $.
## Step 3: Use the point-slope form
Using the point $ (3, -5) $ and $ m = -\frac{7}{3} $ in the point-slope form $ y - y_1 = m(x - x_1) $:
$ y - (-5) = -\frac{7}{3}(x - 3) $
Simplify: $ y + 5 = -\frac{7}{3}x + 7 $
$ y = -\frac{7}{3}x + 2 $

# Answer:
a. The equation is $ y = -\frac{7}{3}x + 2 $ (or in point-slope form: $ y + 5 = -\frac{7}{3}(x - 3) $)

### Problem 7:
# Explanation:
## Step 1: Identify the slope of the given line
The given line is $ y + 9 = -\frac{7}{4}(x - 6) $, so its slope $ m_1 = -\frac{7}{4} $.
## Step 2: Recall the slope of perpendicular lines
The slope of a line perpendicular to a line with slope $ m_1 $ is $ m_2 = \frac{4}{7} $ (negative reciprocal: $ m_2 = -\frac{1}{m_1} = \frac{4}{7} $).
## Step 3: Use the point-slope form
Using the point $ (7, 8) $ and $ m = \frac{4}{7} $ in the point-slope form $ y - y_1 = m(x - x_1) $:
$ y - 8 = \frac{4}{7}(x - 7) $
Simplify: $ y - 8 = \frac{4}{7}x - 4 $
$ y = \frac{4}{7}x + 4 $

# Answer:
a. The equation is $ y = \frac{4}{7}x + 4 $ (or in point-slope form: $ y - 8 = \frac{4}{7}(x - 7) $)

### Problem 8:
# Explanation:
## Step 1: Find the slope of the given line
Rewrite $ 9x + 3y = 6 $ in slope-intercept form ($ y = mx + b $):
$ 3y = -9x + 6 $
$ y = -3x + 2 $
So the slope $ m = -3 $.
## Step 2: Recall the slope of parallel lines
Parallel lines have the same slope, so the slope of the new line is also $ -3 $.
## Step 3: Use the point-slope form
Using the point $ (7, 4) $ and $ m = -3 $ in the point-slope form $ y - y_1 = m(x - x_1) $:
$ y - 4 = -3(x - 7) $
Simplify: $ y - 4 = -3x + 21 $
$ y = -3x + 25 $

# Answer:
a. The equation is $ y = -3x + 25 $ (or in point-slope form: $ y - 4 = -3(x - 7) $)

### Problem 9:
# Explanation:
## Step 1: Find the slope of the given line
Rewrite $ 10x + 3y = -15 $ in slope-intercept form:
$ 3y = -10x - 15 $
$ y = -\frac{10}{3}x - 5 $
So the slope $ m_1 = -\frac{10}{3} $.
## Step 2: Recall the slope of perpendicular lines
The slope of a line perpendicular to a line with slope $ m_1 $ is $ m_2 = \frac{3}{10} $ (negative reciprocal: $ m_2 = -\frac{1}{m_1} = \frac{3}{10} $).
## Step 3: Use the point-slope form
Using the point $ (0, 8) $ (which is the $ y $-intercept, $ b = 8 $) and $ m = \frac{3}{10} $ in the slope-intercept form $ y = mx + b $:
$ y = \frac{3}{10}x + 8 $

# Answer:
a. The equation is $ y = \frac{3}{10}x + 8 $ (or in point-slope form: $ y - 8 = \frac{3}{10}(x - 0) $)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1:

Step 1: Understand the line \( y = 2 \)

Step 2: Find points on the line

Step 3: Determine if it's horizontal or vertical

Step 1: Recall the slope of parallel lines

Step 2: Use the point-slope form

Step 1: Recall the slope of perpendicular lines

Step 2: Determine the equation of the vertical line

Answer:

Problem 2: