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### Problem 19: **Find the pair of perpendicular lines**  
Recall that two lines are **perpendicular** if the product of their slopes is \(-1\).  
The general form of a line is \(Ax + By + C = 0\), and its slope can be found as \(m = -\frac{A}{B}\).

- **(a)**  
  1. Line 1: \(3x - 2y + 4 = 0\)  
     Slope: \(m_1 = -\frac{3}{-2} = \frac{3}{2}\)
  2. Line 2: \(3x + y = 3\) → Rewrite as \(y = -3x + 3\)  
     Slope: \(m_2 = -3\)

  The product \(m_1 \cdot m_2 = \frac{3}{2} \cdot (-3) = -\frac{9}{2} \neq -1\), so not perpendicular.

- **(b)**  
  1. Line 1: \(x - 2y + 5 = 0\) → Slope: \(m_1 = \frac{1}{2}\)
  2. Line 2: \(2x + y = 4\) → Rewrite as \(y = -2x + 4\) → Slope: \(m_2 = -2\)

  The product \(m_1 \cdot m_2 = \frac{1}{2} \cdot (-2) = -1\), so these lines are perpendicular.

- **(c)**  
  1. Line 1: \(-4y + 7x = 0\) → Rewrite as \(y = \frac{7}{4}x\)  
     Slope: \(m_1 = \frac{7}{4}\)
  2. Line 2: \(4x + y = -8\) → Rewrite as \(y = -4x - 8\)  
     Slope: \(m_2 = -4\)

  The product \(m_1 \cdot m_2 = \frac{7}{4} \cdot (-4) = -7 \neq -1\), so not perpendicular.

- **(d)**  
  1. Line 1: \(5x + 3y - 6 = 0\) → Slope: \(m_1 = -\frac{5}{3}\)
  2. Line 2: \(5y - 3x = -15\) → Rewrite as \(y = \frac{3}{5}x - 3\)  
     Slope: \(m_2 = \frac{3}{5}\)

  The product \(m_1 \cdot m_2 = -\frac{5}{3} \cdot \frac{3}{5} = -1\), so these lines are perpendicular.

### Answer for Problem 19:  
- **(b)** and **(d)** are the perpendicular pairs.

---

### Problem 20: **Find the equation of the line parallel to the given one passing through a point**  
For parallel lines, they have the same slope. Let's solve each subproblem by finding the slope of the given line and then writing the equation of the new line using the point-slope form:  
\[ y - y_1 = m(x - x_1) \]

- **(a)**  
  Line: \(y = 5x - 4\) → Slope \(m = 5\).  
  Through point \(A(3, -1)\), the equation is:
  \[
  y - (-1) = 5(x - 3) \Rightarrow y + 1 = 5(x - 3) \Rightarrow y = 5x - 16
  \]

- **(b)**  
  Line: \(-3y + 2x + 8 = 0\) → Rewrite as \(y = \frac{2}{3}x + \frac{8}{3}\) → Slope \(m = \frac{2}{3}\).  
  Through point \(A(0, 0)\), the equation is:
  \[
  y - 0 = \frac{2}{3}(x - 0) \Rightarrow y = \frac{2}{3}x
  \]

- **(c)**  
  Line: \(0.3y - 4x = 9\) → Rewrite as \(y = \frac{40}{3}x + 30\) → Slope \(m = \frac{40}{3}\).  
  Through point \(A(2, 4)\), the equation is:
  \[
  y - 4 = \frac{40}{3}(x - 2)
  \]

- **(d)**  
  Line: \(\frac{1}{2}y - x + 7 = 0\) → Rewrite as \(y = 2x - 14\) → Slope \(m = 2\).  
  Through point \(A(1, 0)\), the equation is:
  \[
  y - 0 = 2(x - 1) \Rightarrow y = 2x - 2
  \]

- **(e)**  
  Line: \(4y - \frac{2}{3}x = 1\) → Rewrite as \(y = \frac{1}{6}x + \frac{1}{4}\) → Slope \(m = \frac{1}{6}\).  
  Through point \(A(-3, 4)\), the equation is:
  \[
  y - 4 = \frac{1}{6}(x + 3)
  \]

### Do you have any questions or need more clarification on these steps?

Here are 5 related questions to deepen your understanding:
1. How do you determine the slope of a line from its general equation?
2. Why is the product of slopes equal to \(-1\) for perpendicular lines?
3. How do you write the equation of a line parallel to another through a given point?
4. Can two lines with the same slope intersect?
5. How would the procedure differ if the lines were neither parallel nor perpendicular?

**Tip:** For quick identification, slopes are easily derived from linear equations by rearranging them into slope-intercept form: \(y = mx + b\).

19. Find the pair of perpendicular lines. 20. Write the equation of the line parallel to the given one passing through the given points.

Learn how to find perpendicular and parallel lines by calculating slopes from linear equations. Solve problems involving lines passing through specific points using point-slope form. Suitable for Grades 9-11 students.

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