Question

2 - 4 practice slopes of parallel and perpendicular lines
for exercises 1 and 2, are the two lines parallel? explain.
for exercises 3 and 4, are the two lines perpendicular? explain.
write an equation of the line parallel and an equation of a line perpendicular to the given line that contains c.

1. c(5, - 2); y=-5x + 3
2. c(8, 1); y = 2x+6
3. c(4, 3); y=-\frac{1}{4}x + 3
4. c(0, 6); y=\frac{2}{3}x + 3

Explanation:

Step1: Recall slope - parallel and perpendicular rules

Parallel lines have equal slopes ($m_1 = m_2$), and for perpendicular lines, the product of their slopes is - 1 ($m_1\times m_2=-1$). The slope - intercept form of a line is $y = mx + b$, where $m$ is the slope.

Step2: For Exercises 1 and 2 (check parallel lines)

Find the slopes of the two lines using the formula $m=\frac{y_2 - y_1}{x_2 - x_1}$ for lines given by two points $(x_1,y_1)$ and $(x_2,y_2)$. If the slopes are equal, the lines are parallel.

Step3: For Exercises 3 and 4 (check perpendicular lines)

Find the slopes of the two lines using $m=\frac{y_2 - y_1}{x_2 - x_1}$. Then check if the product of the slopes is - 1. If it is, the lines are perpendicular.

Step4: For Exercises 5 - 8 (write parallel and perpendicular lines)

For a line $y = mx + b$ and a point $C(x_0,y_0)$:

• Parallel line: The slope of the parallel line is the same as the slope of the given line ($m_{parallel}=m$). Use the point - slope form $y - y_0=m(x - x_0)$ to write the equation of the parallel line.
• Perpendicular line: The slope of the perpendicular line $m_{perpendicular}=-\frac{1}{m}$ (when $m

eq0$). Then use the point - slope form $y - y_0=m_{perpendicular}(x - x_0)$ to write the equation of the perpendicular line.

Let's solve Exercise 5 as an example:
The given line is $y=-5x + 3$, so its slope $m=-5$.

Parallel line:

The slope of the parallel line $m_{parallel}=-5$. The point $C(5,-2)$. Using the point - slope form $y - y_0=m(x - x_0)$, we have $y+2=-5(x - 5)$. Expanding gives $y+2=-5x + 25$, or $y=-5x+23$.

Perpendicular line:

The slope of the perpendicular line $m_{perpendicular}=\frac{1}{5}$ (since $m=-5$ and $m_{perpendicular}=-\frac{1}{m}$). Using the point - slope form with the point $C(5,-2)$, we have $y + 2=\frac{1}{5}(x - 5)$. Expanding gives $y+2=\frac{1}{5}x-1$, or $y=\frac{1}{5}x-3$.

Answer:

For Exercise 5:
Parallel line: $y=-5x + 23$
Perpendicular line: $y=\frac{1}{5}x-3$

(You can follow the same steps to solve the remaining exercises.)