McGraw Hill Glencoe Algebra 1, 2012
MH
McGraw Hill Glencoe Algebra 1, 2012 View details
4. Parallel and Perpendicular Lines
Continue to next subchapter

Exercise 47 Page 244

The slopes m_1 and m_2 of two perpendicular lines satisfy the equation m_1 * m_2 = - 1.

Answer: Carmen is correct.
Explanation: See solution.

Practice makes perfect

To answer this question, we will find the equation of the requested line by ourselves and compare it to the answers Carmen and Chaise obtained. To do so, we will use the point-slope form of a line. y- y_1= m(x- x_1) Here, m is the slope and ( x_1, y_1) is a point on the line. Let's start by finding the slope.

Finding the Slope

If two lines are perpendicular, their slopes m_1 and m_2 are opposite reciprocals. m_1 * m_2 = - 1 Since the given line is written in slope-intercept form, we can identify its slope. y=1/3x+2 Knowing that the slope of the given line is m_1=13, we will use the opposite reciprocal equation to find the value of m_2, the slope of the requested line.
m_1 * m_2 = - 1
1/3 * m_2 = - 1
m_2/3 = - 1
m_2=- 3
The slope of the perpendicular line is - 3. We can substitute it into the point-slope form to obtain a partial equation. y-y_1=- 3(x-x_1)

Completing the Equation

Next, we will use the fact that the line passes through the point ( - 3, 5) to complete our equation.
y-y_1=- 3(x-x_1)
y- 5=- 3( x-( - 3))
â–Ľ
Solve for y
y-5=- 3(x+3)
y-5=- 3x-9
y=- 3x-4
The equation of the perpendicular line to y= 13x+2 through (- 3,5) is y=- 3x-4. Therefore, Carmen is correct. Note that Chase's error was using the incorrect slope, he used 3 instead of - 3.