https://file.aiquickdraw.com/mathbot/file/1725826426884q1tbao8m.jpg

Let's break down the problem and solve each part step by step.

### **Given:**
- Line \( l \) has the equation \( 4x - 3y + 8 = 0 \) and passes through point \( B(7, 12) \).
- Line \( m \) is perpendicular to line \( l \) and passes through point \( B \).

### **Part (a): Find the equation of line m.**
1. **Equation of line \( l \):**  
   We know the equation of line \( l \) is \( 4x - 3y + 8 = 0 \). The slope of this line can be found by converting the equation into slope-intercept form \( y = mx + b \).

   \[
   4x - 3y + 8 = 0 \implies 3y = 4x + 8 \implies y = \frac{4}{3}x + \frac{8}{3}
   \]
   So, the slope \( m_l \) of line \( l \) is \( \frac{4}{3} \).

2. **Perpendicular slope:**  
   The slope of a line perpendicular to line \( l \) will be the negative reciprocal of \( \frac{4}{3} \). Therefore, the slope of line \( m \), \( m_m \), is:
   \[
   m_m = -\frac{3}{4}
   \]

3. **Equation of line \( m \):**  
   Line \( m \) passes through point \( B(7, 12) \) and has slope \( -\frac{3}{4} \). Using the point-slope form \( y - y_1 = m(x - x_1) \):
   \[
   y - 12 = -\frac{3}{4}(x - 7)
   \]
   Expanding this:
   \[
   y - 12 = -\frac{3}{4}x + \frac{21}{4}
   \]
   Multiply the whole equation by 4 to eliminate the fraction:
   \[
   4(y - 12) = -3(x - 7)
   \]
   \[
   4y - 48 = -3x + 21
   \]
   Bringing everything to one side:
   \[
   3x + 4y - 69 = 0
   \]
   So, the equation of line \( m \) is:
   \[
   3x + 4y - 69 = 0
   \]

### **Part (b): Find the coordinates of points A and C.**

- **Point A (where line \( l \) crosses the x-axis):**
  On the x-axis, \( y = 0 \). Substitute \( y = 0 \) in the equation of line \( l \):
  \[
  4x - 3(0) + 8 = 0 \implies 4x + 8 = 0 \implies x = -2
  \]
  So, point \( A \) is \( (-2, 0) \).

- **Point C (where line \( m \) crosses the x-axis):**
  Similarly, for line \( m \), substitute \( y = 0 \) in the equation \( 3x + 4y - 69 = 0 \):
  \[
  3x + 4(0) - 69 = 0 \implies 3x = 69 \implies x = 23
  \]
  So, point \( C \) is \( (23, 0) \).

### **Part (c): Find the area of triangle ABC.**

We have the coordinates of points \( A(-2, 0) \), \( B(7, 12) \), and \( C(23, 0) \). The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates \( (-2, 0) \), \( (7, 12) \), and \( (23, 0) \):
\[
\text{Area} = \frac{1}{2} \left| (-2)(12 - 0) + 7(0 - 0) + 23(0 - 12) \right|
\]
Simplifying:
\[
\text{Area} = \frac{1}{2} \left| (-2)(12) + 7(0) + 23(-12) \right|
\]
\[
\text{Area} = \frac{1}{2} \left| -24 - 276 \right| = \frac{1}{2} \times 300 = 150
\]
So, the area of triangle \( ABC \) is \( \boxed{150} \).

### **Summary of Solutions:**
- (a) The equation of line \( m \) is \( 3x + 4y - 69 = 0 \).
- (b) The coordinates of points \( A \) and \( C \) are \( A(-2, 0) \) and \( C(23, 0) \).
- (c) The area of triangle \( ABC \) is \( 150 \).

Do you want further clarification on any step or additional details?

### **Relative Questions:**
1. How do you derive the slope of a line from its equation?
2. What is the significance of perpendicular slopes in geometry?
3. Can the area of a triangle be negative? Why or why not?
4. How would the problem change if line \( m \) wasn't perpendicular to line \( l \)?
5. How do you interpret the equation of a line when expressed in general form \( ax + by + c = 0 \)?

### **Tip:**  
When two lines are perpendicular, their slopes multiply to give \( -1 \). This fact is very useful for finding equations of perpendicular lines.

Line l has equation 4x - 3y + 8 = 0 and passes through point B (7, 12). Line m is perpendicular to line l and passes through point B. (a) Find the equation of line m in the form ax + by + c = 0 where a, b and c are integers. (b) Find the coordinates of points A and C. (c) Find the area of triangle ABC.

Learn how to find the equation of a perpendicular line to 4x - 3y + 8 = 0 passing through point B(7, 12) and calculate the area of triangle ABC. This problem covers key geometry concepts including slopes, perpendicular lines, and the area of a triangle. Suitable for Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation