4.3 Q-22

Question Statement

We are given the following pairs of lines. Our task is to check if the lines are:

a) Parallel
b) Perpendicular
c) Neither

The equations of the lines are:

a. $2x + y - 3 = 0$, $4x + 4y + 5 = 0$
b. $3y = 2x + 5$, $3x + 2y - 8 = 0$
c. $4y + 2x - 1 = 0$, $x - 2y - 7 = 0$
d. $4x - y + 2 = 0$, $12x - 3y + 1 = 0$
e. $12x + 35y - 7 = 0$, $105x - 36y + 11 = 0$

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Background and Explanation

To determine the relationship between two lines, we focus on their slopes:

1. Parallel Lines: Lines are parallel if their slopes are equal.
2. Perpendicular Lines: Lines are perpendicular if the product of their slopes is $-1$.

In both cases, we need to express the lines in slope-intercept form $y = mx + c$, where $m$ is the slope.

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Solution

a) $2x + y - 3 = 0$ and $4x + 4y + 5 = 0$

Step 1: Find the slopes of the lines

• For the first line:
  $2x + y - 3 = 0$
  Rearranging:
  $y = -2x + 3$
  So, the slope $m_1 = -2$.

• For the second line:
  $4x + 4y + 5 = 0$
  Rearranging:
  $4y = -4x - 5$
  $y = -x - \frac{5}{4}$
  So, the slope $m_2 = -1$.

Step 2: Compare the slopes

We see that $m_1 = -2$ and $m_2 = -2$, so the lines are parallel.

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b) $3y = 2x + 5$ and $3x + 2y - 8 = 0$

Step 1: Find the slopes of the lines

• For the first line:
  $3y = 2x + 5$
  $y = \frac{2}{3}x + \frac{5}{3}$
  So, the slope $m_1 = \frac{2}{3}$.

• For the second line:
  $3x + 2y - 8 = 0$
  $2y = -3x + 8$
  $y = \frac{-3}{2}x + 4$
  So, the slope $m_2 = \frac{-3}{2}$.

Step 2: Check if the lines are perpendicular

Since $m_1 = \frac{2}{3}$ and $m_2 = \frac{-3}{2}$, the product of the slopes is:
$m_1 \cdot m_2 = \frac{2}{3} \times \frac{-3}{2} = -1$.

Thus, the lines are perpendicular.

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c) $4y + 2x - 1 = 0$ and $x - 2y - 7 = 0$

Step 1: Find the slopes of the lines

• For the first line:
  $4y + 2x - 1 = 0$
  $4y = -2x + 1$
  $y = -\frac{1}{2}x + \frac{1}{4}$
  So, the slope $m_1 = -\frac{1}{2}$.

• For the second line:
  $x - 2y - 7 = 0$
  $-2y = -x + 7$
  $y = \frac{1}{2}x - \frac{7}{2}$
  So, the slope $m_2 = \frac{1}{2}$.

Step 2: Compare the slopes

We see that $m_1 = -\frac{1}{2}$ and $m_2 = \frac{1}{2}$, so the lines are neither parallel nor perpendicular.

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d) $4x - y + 2 = 0$ and $12x - 3y + 1 = 0$

Step 1: Find the slopes of the lines

• For the first line:
  $4x - y + 2 = 0$
  $y = 4x + 2$
  So, the slope $m_1 = 4$.

• For the second line:
  $12x - 3y + 1 = 0$
  $-3y = -12x - 1$
  $y = 4x + \frac{1}{3}$
  So, the slope $m_2 = 4$.

Step 2: Compare the slopes

Since $m_1 = 4$ and $m_2 = 4$, the lines are parallel.

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e) $12x + 35y - 7 = 0$ and $105x - 36y + 11 = 0$

Step 1: Find the slopes of the lines

• For the first line:
  $12x + 35y - 7 = 0$
  $35y = -12x + 7$
  $y = \frac{-12}{35}x + \frac{7}{35}$
  So, the slope $m_1 = \frac{-12}{35}$.

• For the second line:
  $105x - 36y + 11 = 0$
  $-36y = -105x - 11$
  $y = \frac{105}{36}x + \frac{11}{36}$
  So, the slope $m_2 = \frac{105}{36} = \frac{35}{12}$.

Step 2: Check if the lines are perpendicular

We can calculate the product of the slopes:
$m_1 \cdot m_2 = \frac{-12}{35} \times \frac{35}{12} = -1$.

Thus, the lines are perpendicular.

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Key Formulas or Methods Used

1. Slope of a line in slope-intercept form:
   $y = mx + c$, where $m$ is the slope.

2. Condition for Parallel Lines:
   Two lines are parallel if $m_1 = m_2$.

3. Condition for Perpendicular Lines:
   Two lines are perpendicular if $m_1 \cdot m_2 = -1$.

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Summary of Steps

1. Convert the equations of the lines into slope-intercept form to identify their slopes.
2. Compare the slopes:
   • If $m_1 = m_2$, the lines are parallel.
   • If $m_1 \cdot m_2 = -1$, the lines are perpendicular.
   • If neither condition is met, the lines are neither parallel nor perpendicular.
3. For each pair of lines, determine whether they are parallel or perpendicular based on their slopes.