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Math  /  Algebra

QuestionName \qquad Period \qquad Date \qquad Module 5 Test (ALG 1) - MODIFIED DIRECTIONS: Read and answer the following questions. Make sure to show your work and write legibly.
1. Which equation written in point-slope form represents a line passing through the point (4,−8)(4,-8) with a slope of −14?y−y1=m(x−x1)-\frac{1}{4} ? \quad \boldsymbol{y}-y_{1}=m\left(x-x_{1}\right) A. y+8=−14(x+4)y+8=-\frac{1}{4}(x+4)
18. y+8=−14(x−4)y+8=-\frac{1}{4}(x-4) C. y=8=−14(x+4)y=8=-\frac{1}{4}(x+4) D. y−8=−14(x−4)y-8=-\frac{1}{4}(x-4)
2. Which equation models the line on the graph? y−y1=m(x−x1)\boldsymbol{y}-y_{1}=m\left(\boldsymbol{x}-x_{1}\right) A. y−2=−2(x−3)y-2=-2(x-3) B. y−2=−12(x−3)y-2=-\frac{1}{2}(x-3) C. y+2=−12(x+3)y+2=-\frac{1}{2}(x+3) D. y+2=−2(x+3)y+2=-2(x+3)
4. Write an equation in slope-intercept form with a slope of 3 and passes through the point (−2,−1)(-2,-1). Step 1: y−=(x−x1)\boldsymbol{y}-\quad=\left(\boldsymbol{x}-x_{1}\right) Step 2: Distribute Step 3: Isolate y A. y=3x+7y=3 x+7 B. y=3x+6y=3 x+6 C. y=3x+5y=3 x+5 D. y=3x−1y=3 x-1
8. Which equation is parallel to the line y=−4x−8?y=-4 x-8 ? A. y=14x−8y=\frac{1}{4} x-8 B. y=4x−8y=4 x-8 C. y=−4x+5y=-4 x+5 D. y=−14x+3y=-\frac{1}{4} x+3
6. Write the following equation in slope-intercept form. y−4=6(x−2)y-4=6(x-2)
9. Find the equation of a line that is perpendicular to the line represented y=−13x+4y=-\frac{1}{3} x+4. A. y=13x+3y=\frac{1}{3} x+3 B. y=−13x−3y=-\frac{1}{3} x-3 C. y=3x+13y=3 x+13 D. y=−3x−5y=-3 x-5 24−2524-25

Studdy Solution

STEP 1

1. We are given various problems related to equations of lines, including point-slope form, slope-intercept form, parallel and perpendicular lines.
2. We need to apply the formulas for point-slope and slope-intercept forms to solve these problems.
3. The slope-intercept form of a line is y=mx+b y = mx + b .
4. Lines that are parallel have the same slope.
5. Lines that are perpendicular have slopes that are negative reciprocals of each other.

STEP 2

1. Solve problem 1: Identify the correct point-slope form equation.
2. Solve problem 2: Identify the correct point-slope form equation from the graph.
3. Solve problem 4: Write an equation in slope-intercept form.
4. Solve problem 8: Identify the equation parallel to a given line.
5. Solve problem 6: Convert an equation to slope-intercept form.
6. Solve problem 9: Find the equation of a line perpendicular to a given line.

STEP 3

Identify the point-slope form equation that represents a line passing through the point (4,−8)(4, -8) with a slope of −14-\frac{1}{4}.
The point-slope form is given by: y−y1=m(x−x1) y - y_1 = m(x - x_1)
Substitute m=−14 m = -\frac{1}{4} , x1=4 x_1 = 4 , and y1=−8 y_1 = -8 : y−(−8)=−14(x−4) y - (-8) = -\frac{1}{4}(x - 4) y+8=−14(x−4) y + 8 = -\frac{1}{4}(x - 4)
The correct answer is B: y+8=−14(x−4) y + 8 = -\frac{1}{4}(x - 4) .

STEP 4

Identify the correct point-slope form equation from the graph.
Assuming the graph shows a line passing through the point (3,2)(3, 2) with a slope of −2-2, the point-slope form is: y−2=−2(x−3) y - 2 = -2(x - 3)
The correct answer is A: y−2=−2(x−3) y - 2 = -2(x - 3) .

STEP 5

Write an equation in slope-intercept form with a slope of 3 that passes through the point (−2,−1)(-2, -1).
Start with the point-slope form: y−(−1)=3(x−(−2)) y - (-1) = 3(x - (-2)) y+1=3(x+2) y + 1 = 3(x + 2)
Distribute: y+1=3x+6 y + 1 = 3x + 6
Isolate y y : y=3x+6−1 y = 3x + 6 - 1 y=3x+5 y = 3x + 5
The correct answer is C: y=3x+5 y = 3x + 5 .

STEP 6

Identify the equation parallel to the line y=−4x−8 y = -4x - 8 .
Parallel lines have the same slope. The slope of the given line is −4-4.
The correct answer is C: y=−4x+5 y = -4x + 5 .

STEP 7

Convert the equation y−4=6(x−2) y - 4 = 6(x - 2) to slope-intercept form.
Start with the given equation: y−4=6(x−2) y - 4 = 6(x - 2)
Distribute: y−4=6x−12 y - 4 = 6x - 12
Isolate y y : y=6x−12+4 y = 6x - 12 + 4 y=6x−8 y = 6x - 8

STEP 8

Find the equation of a line that is perpendicular to the line represented by y=−13x+4 y = -\frac{1}{3}x + 4 .
Perpendicular lines have slopes that are negative reciprocals. The slope of the given line is −13-\frac{1}{3}, so the perpendicular slope is 33.
The correct answer is C: y=3x+13 y = 3x + 13 .

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