Question

1. determine if the lines are parallel, perpendicular, or neither.

a. $y = \frac{1}{4}x + 5$\quad\quad b. $-6y - 10x = 14$\quad\quad c. $y = -\frac{3}{2}x + 7$
\quad\quad $4y - 20 = x$\quad\quad\quad\quad $5y = 32 + 7x$\quad\quad\quad\quad $y = \frac{2}{3}x + 15$

Explanation:

Step1: Analyze part a

First, rewrite the second equation \(4y - 20=x\) in slope - intercept form (\(y = mx + b\), where \(m\) is the slope).
Solve \(4y-20 = x\) for \(y\):
Add 20 to both sides: \(4y=x + 20\)
Divide both sides by 4: \(y=\frac{1}{4}x + 5\)
The first equation is \(y=\frac{1}{4}x + 5\). Since the slopes of the two lines are both \(\frac{1}{4}\) (and the y - intercepts are also the same, but for parallel lines, we only need equal slopes), the two lines are parallel.

Step2: Analyze part b

Rewrite the first equation \(-6y-10x = 14\) in slope - intercept form.
Add \(10x\) to both sides: \(-6y=10x + 14\)
Divide both sides by \(-6\): \(y=-\frac{10}{6}x-\frac{14}{6}=-\frac{5}{3}x-\frac{7}{3}\)
Rewrite the second equation \(5y = 32+7x\) in slope - intercept form.
Divide both sides by 5: \(y=\frac{7}{5}x+\frac{32}{5}\)
The slope of the first line is \(m_1 =-\frac{5}{3}\) and the slope of the second line is \(m_2=\frac{7}{5}\).
Check if they are parallel: \(-\frac{5}{3}
eq\frac{7}{5}\), so not parallel.
Check if they are perpendicular: The product of the slopes \(m_1\times m_2=-\frac{5}{3}\times\frac{7}{5}=-\frac{7}{3}
eq - 1\), so neither parallel nor perpendicular.

Step3: Analyze part c

The first line is \(y =-\frac{3}{2}x + 7\) with slope \(m_1=-\frac{3}{2}\)
The second line is \(y=\frac{2}{3}x + 15\) with slope \(m_2=\frac{2}{3}\)
Check the product of the slopes: \(m_1\times m_2=-\frac{3}{2}\times\frac{2}{3}=- 1\). So the two lines are perpendicular.

Answer:

a. The two lines are parallel.
b. The two lines are neither parallel nor perpendicular.
c. The two lines are perpendicular.