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

Explanation:

Step1: Recall slope - formula

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$, where $(x_1,y_1)$ and $(x_2,y_2)$ are two points on the line.

Step2: For parallel lines

Two lines are parallel if and only if their slopes are equal.

Step3: For perpendicular lines

Two lines are perpendicular if and only if the product of their slopes is - 1, i.e., $m_1\times m_2=-1$.

Step4: Solve Exercise 1

Let's find the slope of line $\ell_1$ with points $(-3,0)$ and $(3,2)$. Using the slope formula $m_1=\frac{2 - 0}{3-(-3)}=\frac{2}{6}=\frac{1}{3}$. For line $\ell_2$ with points $(-1,-3)$ and $(5,-1)$, $m_2=\frac{-1-(-3)}{5 - (-1)}=\frac{2}{6}=\frac{1}{3}$. Since $m_1 = m_2=\frac{1}{3}$, the two lines are parallel.

Step5: Solve Exercise 2

For line $\ell_1$ with points $(-3,6)$ and $(-1,-3)$, $m_1=\frac{-3 - 6}{-1-(-3)}=\frac{-9}{2}$. For line $\ell_2$ with points $(4,8)$ and $(6,0)$, $m_2=\frac{0 - 8}{6 - 4}=\frac{-8}{2}=-4$. Since $m_1
eq m_2$, the two lines are not parallel.

Step6: Solve Exercise 3

For line $\ell_1$ with points $(-2,-1)$ and $(1,-4)$, $m_1=\frac{-4-(-1)}{1-(-2)}=\frac{-3}{3}=-1$. For line $\ell_2$ with points $(-2,-8)$ and $(5,-3)$, $m_2=\frac{-3-(-8)}{5-(-2)}=\frac{5}{7}$. Since $m_1\times m_2=(-1)\times\frac{5}{7}=-\frac{5}{7}
eq - 1$, the two lines are not perpendicular.

Step7: Solve Exercise 4

For line $\ell_1$ with points $(-5,6)$ and $(-1,-2)$, $m_1=\frac{-2 - 6}{-1-(-5)}=\frac{-8}{4}=-2$. For line $\ell_2$ with points $(-5,0)$ and $(1,3)$, $m_2=\frac{3 - 0}{1-(-5)}=\frac{3}{6}=\frac{1}{2}$. Since $m_1\times m_2=(-2)\times\frac{1}{2}=-1$, the two lines are perpendicular.

Step8: Solve Exercise 5

The given line is $y=-5x + 3$, so its slope $m=-5$.
A line parallel to it passing through $C(5,-2)$ has the same slope $m=-5$. Using the point - slope form $y - y_1=m(x - x_1)$, we have $y-(-2)=-5(x - 5)$, which simplifies to $y+2=-5x + 25$, or $y=-5x+23$.
A line perpendicular to it has a slope $m'=\frac{1}{5}$ (since $-5\times m'=-1$). Using the point - slope form with $(x_1,y_1)=(5,-2)$, we get $y-(-2)=\frac{1}{5}(x - 5)$, which simplifies to $y + 2=\frac{1}{5}x-1$, or $y=\frac{1}{5}x-3$.

Step9: Solve Exercise 6

The given line is $y = 2x+6$, so its slope $m = 2$.
A line parallel to it passing through $C(8,1)$ has the same slope $m = 2$. Using the point - slope form $y - y_1=m(x - x_1)$, we have $y - 1=2(x - 8)$, which simplifies to $y-1=2x-16$, or $y=2x-15$.
A line perpendicular to it has a slope $m'=-\frac{1}{2}$ (since $2\times m'=-1$). Using the point - slope form with $(x_1,y_1)=(8,1)$, we get $y - 1=-\frac{1}{2}(x - 8)$, which simplifies to $y-1=-\frac{1}{2}x + 4$, or $y=-\frac{1}{2}x+5$.

Answer:

Exercise 1: The two lines are parallel.
Exercise 2: The two lines are not parallel.
Exercise 3: The two lines are not perpendicular.
Exercise 4: The two lines are perpendicular.
Exercise 5: Parallel line: $y=-5x + 23$, Perpendicular line: $y=\frac{1}{5}x-3$.
Exercise 6: Parallel line: $y=2x-15$, Perpendicular line: $y=-\frac{1}{2}x + 5$.