Question

directions: determine if the given equations are parallel, perpendicular, or neither.

1. $3x + 2y = 6$ and $y = -\frac{3}{2}x + 5$
2. $3y = 4x + 15$ and $9x + 12y = 12$
3. $8x - 2y = 4$ and $x + 4y = -12$
4. $3x + 2y = 10$ and $2x + 3y = -3$
5. $-4y = -2x + 8$ and $3x - 6y = 6$
6. $y = 8$ and $x = -1$

Explanation:

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Problem 7: $3x + 2y = 6$ and $y = -\frac{3}{2}x + 5$

Step1: Isolate $y$ for first equation

$3x + 2y = 6 \implies 2y = -3x + 6 \implies y = -\frac{3}{2}x + 3$

Step2: Compare slopes

Slope of line 1: $m_1 = -\frac{3}{2}$; Slope of line 2: $m_2 = -\frac{3}{2}$. Parallel lines have equal slopes.
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Problem 8: $3y = 4x + 15$ and $9x + 12y = 12$

Step1: Isolate $y$ for first equation

$3y = 4x + 15 \implies y = \frac{4}{3}x + 5$

Step2: Isolate $y$ for second equation

$9x + 12y = 12 \implies 12y = -9x + 12 \implies y = -\frac{3}{4}x + 1$

Step3: Check slope relationship

$m_1 = \frac{4}{3}$, $m_2 = -\frac{3}{4}$. $m_1 \times m_2 = \frac{4}{3} \times -\frac{3}{4} = -1$. Perpendicular lines have slopes whose product is -1.
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Problem 9: $8x - 2y = 4$ and $x + 4y = -12$

Step1: Isolate $y$ for first equation

$8x - 2y = 4 \implies -2y = -8x + 4 \implies y = 4x - 2$

Step2: Isolate $y$ for second equation

$x + 4y = -12 \implies 4y = -x -12 \implies y = -\frac{1}{4}x - 3$

Step3: Check slope relationship

$m_1 = 4$, $m_2 = -\frac{1}{4}$. $m_1 \times m_2 = 4 \times -\frac{1}{4} = -1$.
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Problem 10: $3x + 2y = 10$ and $2x + 3y = -3$

Step1: Isolate $y$ for first equation

$3x + 2y = 10 \implies 2y = -3x + 10 \implies y = -\frac{3}{2}x + 5$

Step2: Isolate $y$ for second equation

$2x + 3y = -3 \implies 3y = -2x -3 \implies y = -\frac{2}{3}x - 1$

Step3: Compare slopes

$m_1 = -\frac{3}{2}$, $m_2 = -\frac{2}{3}$. Slopes are not equal, and their product is $(-\frac{3}{2}) \times (-\frac{2}{3}) = 1
eq -1$.
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Problem 11: $-4y = -2x + 8$ and $3x - 6y = 6$

Step1: Isolate $y$ for first equation

$-4y = -2x + 8 \implies y = \frac{1}{2}x - 2$

Step2: Isolate $y$ for second equation

$3x - 6y = 6 \implies -6y = -3x + 6 \implies y = \frac{1}{2}x - 1$

Step3: Compare slopes

$m_1 = \frac{1}{2}$, $m_2 = \frac{1}{2}$. Slopes are equal.
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Problem 12: $y = 8$ and $x = -1$

Step1: Identify line types

$y=8$ is a horizontal line (slope $m_1=0$); $x=-1$ is a vertical line (undefined slope). Horizontal and vertical lines are perpendicular.
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Answer:

1. Parallel
2. Perpendicular
3. Perpendicular
4. Neither
5. Parallel
6. Perpendicular