Question

directions: write an equation passing through the point that is perpendicular to the given equation. 9. (3, -3); ( y = \frac{3}{4}x + 5 ) 10. (-3, -2); ( y = x - 2 ) 11. (2, 3); ( 2x + 10y = 20 ) 12. (-1, -6); ( x + 3y = 6 )

Explanation:

Problem 9:

Step1: Find slope of perpendicular line

The given line is \( y = \frac{3}{4}x + 5 \), so its slope \( m_1 = \frac{3}{4} \). For perpendicular lines, \( m_2 = -\frac{1}{m_1} = -\frac{4}{3} \).

Step2: Use point - slope form

Point - slope formula: \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1)=(3, - 3) \) and \( m = -\frac{4}{3} \).
Substitute values: \( y - (-3)=-\frac{4}{3}(x - 3) \)
Simplify: \( y + 3=-\frac{4}{3}x + 4 \)
Subtract 3: \( y=-\frac{4}{3}x + 1 \)

Problem 10:

Step1: Find slope of perpendicular line

Given line \( y = x - 2 \), slope \( m_1 = 1 \). Perpendicular slope \( m_2=-1 \) (since \( m_1\times m_2=-1 \)).

Step2: Use point - slope form

Point \( (-3,-2) \), \( m=-1 \). Point - slope: \( y - (-2)=-1(x - (-3)) \)
Simplify: \( y + 2=-x - 3 \)
Subtract 2: \( y=-x - 5 \)

Problem 11:

Step1: Find slope of given line

Rewrite \( 2x + 10y = 20 \) in slope - intercept form (\( y=mx + b \)):
\( 10y=-2x + 20 \)
\( y=-\frac{1}{5}x + 2 \). Slope \( m_1 = -\frac{1}{5} \). Perpendicular slope \( m_2 = 5 \) (since \( m_1\times m_2=-1 \)).

Step2: Use point - slope form

Point \( (2,3) \), \( m = 5 \). Point - slope: \( y - 3 = 5(x - 2) \)
Simplify: \( y - 3=5x - 10 \)
Add 3: \( y = 5x - 7 \)

Problem 12:

Answer:

s:

1. \( \boldsymbol{y = -\frac{4}{3}x + 1} \)
2. \( \boldsymbol{y=-x - 5} \)
3. \( \boldsymbol{y = 5x - 7} \)
4. \( \boldsymbol{y = 3x - 3} \)