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consider the line y=9x−3. find the equation of the line that is perpend…

Question

consider the line y=9x−3. find the equation of the line that is perpendicular to this line and passes through the point (3, 6). find the equation of the line that is parallel to this line and passes through the point (3, 6). note that the aleks graphing calculator may be helpful in checking your answer. equation of perpendicular line: equation of parallel line:

Explanation:

Step1: Find slope of original line

The line \( y = 9x - 3 \) is in slope - intercept form \( y=mx + b \), where \( m \) is the slope. So the slope of the given line, \( m_{original}=9 \).

Step2: Equation of perpendicular line

  1. The slope of a line perpendicular to a line with slope \( m \) is \( m_{perpendicular}=-\frac{1}{m} \). For \( m = 9 \), \( m_{perpendicular}=-\frac{1}{9} \).
  2. We use the point - slope form of a line \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(3,6) \) and \( m =-\frac{1}{9} \).
  • Substitute the values into the point - slope formula: \( y - 6=-\frac{1}{9}(x - 3) \).
  • Expand the right - hand side: \( y - 6=-\frac{1}{9}x+\frac{1}{3} \).
  • Add 6 to both sides: \( y=-\frac{1}{9}x+\frac{1}{3}+6 \).
  • Convert 6 to thirds: \( 6=\frac{18}{3} \), so \( y=-\frac{1}{9}x+\frac{1 + 18}{3}=-\frac{1}{9}x+\frac{19}{3} \).

Step3: Equation of parallel line

  1. The slope of a line parallel to a line with slope \( m \) is equal to \( m \). So \( m_{parallel}=9 \).
  2. Use the point - slope form \( y - y_1=m(x - x_1) \) with \( (x_1,y_1)=(3,6) \) and \( m = 9 \).
  • Substitute the values: \( y - 6=9(x - 3) \).
  • Expand the right - hand side: \( y - 6=9x-27 \).
  • Add 6 to both sides: \( y=9x-27 + 6=9x-21 \).

Answer:

Equation of perpendicular line: \( y=-\frac{1}{9}x+\frac{19}{3} \)
Equation of parallel line: \( y = 9x-21 \)