Question

in exercises 1 - 4, describe the slope of the line. then find the slope. (see example 1.)
in exercises 5 and 6, find the slope of the line that passes through the given points.

1. (1,4),(3, - 6)
2. (2, - 2),( - 7, - 5)

in exercises 7 - 10, the points represented by the table lie on a line. find the slope of the line. (see example 2.)
7.

x	- 9	- 5	- 1	3
y	- 2	0	2	4

8.

x	- 1	2	5	8
y	- 6	- 6	- 6	- 6

9.

x	0	0	0	0
y	- 4	0	4	8

10.

x	- 4	- 3	- 2	- 1
y	2	- 5	- 12	- 19

Explanation:

Step1: Recall slope - formula

The slope $m$ of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Solve Exercise 5

For points $(1,4)$ and $(3, - 6)$, let $(x_1,y_1)=(1,4)$ and $(x_2,y_2)=(3,-6)$. Then $m=\frac{-6 - 4}{3 - 1}=\frac{-10}{2}=-5$.

Step3: Solve Exercise 6

For points $(2,-2)$ and $(-7,-5)$, let $(x_1,y_1)=(2,-2)$ and $(x_2,y_2)=(-7,-5)$. Then $m=\frac{-5+2}{-7 - 2}=\frac{-3}{-9}=\frac{1}{3}$.

Step4: Solve Exercise 7

Take two points, say $(-9,-2)$ and $(-5,0)$. Then $m=\frac{0 + 2}{-5+9}=\frac{2}{4}=\frac{1}{2}$.

Step5: Solve Exercise 8

Since $y=-6$ for all $x$ - values, the line is horizontal. Using the slope formula with any two points, say $(-1,-6)$ and $(2,-6)$, $m=\frac{-6+6}{2 + 1}=0$.

Step6: Solve Exercise 9

The $x$ - values are all $0$, so the line is vertical. The slope of a vertical line is undefined.

Step7: Solve Exercise 10

Take two points, say $(-4,2)$ and $(-3,-5)$. Then $m=\frac{-5 - 2}{-3+4}=\frac{-7}{1}=-7$.

Answer:

Exercise 5: - 5
Exercise 6: $\frac{1}{3}$
Exercise 7: $\frac{1}{2}$
Exercise 8: 0
Exercise 9: Undefined
Exercise 10: - 7