QUESTION IMAGE
Question
perpendicular lines mock
similar to the questions about
parallel lines, on the act this
typically takes the form of being
given an equation in the question
and needing to solve the
equation of a line that is
perpendicular to the given line.
which line below is perpendicular to ( 5x + 6y = 18 )?
a. ( y = -\frac{6}{5}x + 8 )
b. ( y = -\frac{5}{6}x + \frac{6}{5} )
c. ( y = \frac{5}{6}x + 2 )
d. ( y = \frac{5}{6}x + \frac{6}{5} )
e. ( y = \frac{6}{5}x + 3 )
Step1: Find slope of given line
First, rewrite the given line \(5x + 6y = 18\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept).
Solve for \(y\):
\(6y=-5x + 18\)
\(y=-\frac{5}{6}x+3\)
The slope of the given line \(m_1 =-\frac{5}{6}\).
Step2: Find slope of perpendicular line
If two lines are perpendicular, the product of their slopes \(m_1\times m_2=- 1\). Let the slope of the perpendicular line be \(m_2\).
We know \(m_1 =-\frac{5}{6}\), so \(-\frac{5}{6}\times m_2=-1\)
Solve for \(m_2\): \(m_2=\frac{6}{5}\)
Now we check the slopes of the given options:
- Option a: \(y =-\frac{6}{5}x + 8\), slope \(m=-\frac{6}{5}\)
- Option b: \(y=-\frac{5}{6}x+\frac{6}{5}\), slope \(m =-\frac{5}{6}\)
- Option c: \(y=\frac{5}{6}x + 2\), slope \(m=\frac{5}{6}\)
- Option d: \(y=\frac{5}{6}x+\frac{6}{5}\), slope \(m=\frac{5}{6}\)
- Option e: \(y=\frac{6}{5}x + 3\), slope \(m=\frac{6}{5}\)
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e. \(y=\frac{6}{5}x + 3\)