Question

section 2.3 summary
in this section, we discussed graphs and equations of lines. lines are often expressed in two forms:

• general form: $ax + by = c$
• slope - intercept form: $y = mx + b$

all lines (except horizontal and vertical) have exactly one x - intercept and exactly one y - intercept. the slope of a line is a measure of steepness.

• slope of a line passing through $(x_1,y_1)$ and $(x_2,y_2)$: $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{rise}{run}$
• horizontal lines: $m = 0$
• vertical lines: $m$ is undefined

we found equations of lines, given either two points or the slope and a point. we found the point - slope form, $y - y_1=m(x - x_1)$, useful when the slope and a point are given. we also discussed both parallel (nonintersecting) and perpendicular (forming a right angle) lines. parallel lines have the same slope. perpendicular lines have negative reciprocal slopes, provided their slopes are defined.
section 2.3 exercises

• skills

in exercises 1 - 10, find the slope of the line that passes through the given points.

1. $(1,3)$ and $(2,6)$
2. $(2,1)$ and $(4,9)$
3. $(-2,5)$ and $(2,-3)$
4. $(-1,-4)$ and $(4,6)$
5. $(-7,9)$ and $(3,-10)$
6. $(11,-3)$ and $(-2,6)$
7. $(0.2,-1.7)$ and $(3.1,5.2)$
8. $(-2.4,1.7)$ and $(-5.6,-2.3)$
9. $(\frac{2}{3},-\frac{1}{4})$ and $(\frac{5}{6},-\frac{3}{4})$
10. $(\frac{1}{2},\frac{3}{5})$ and $(-\frac{3}{4},\frac{7}{5})$

Explanation:

Step1: Recall slope - formula

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

Step2: Solve for exercise 1

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

Step3: Solve for exercise 2

For points $(2,1)$ and $(4,9)$, let $(x_1,y_1)=(2,1)$ and $(x_2,y_2)=(4,9)$. Then $m=\frac{9 - 1}{4 - 2}=\frac{8}{2}=4$.

Step4: Solve for exercise 3

For points $(-2,5)$ and $(2,-3)$, let $(x_1,y_1)=(-2,5)$ and $(x_2,y_2)=(2,-3)$. Then $m=\frac{-3 - 5}{2-(-2)}=\frac{-8}{4}=-2$.

Step5: Solve for exercise 4

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

Step6: Solve for exercise 5

For points $(-7,9)$ and $(3,-10)$, let $(x_1,y_1)=(-7,9)$ and $(x_2,y_2)=(3,-10)$. Then $m=\frac{-10 - 9}{3-(-7)}=\frac{-19}{10}=-1.9$.

Step7: Solve for exercise 6

For points $(11,-3)$ and $(-2,6)$, let $(x_1,y_1)=(11,-3)$ and $(x_2,y_2)=(-2,6)$. Then $m=\frac{6-(-3)}{-2 - 11}=\frac{9}{-13}=-\frac{9}{13}$.

Step8: Solve for exercise 7

For points $(0.2,-1.7)$ and $(3.1,5.2)$, let $(x_1,y_1)=(0.2,-1.7)$ and $(x_2,y_2)=(3.1,5.2)$. Then $m=\frac{5.2-(-1.7)}{3.1 - 0.2}=\frac{6.9}{2.9}=\frac{69}{29}$.

Step9: Solve for exercise 8

For points $(-2.4,1.7)$ and $(-5.6,-2.3)$, let $(x_1,y_1)=(-2.4,1.7)$ and $(x_2,y_2)=(-5.6,-2.3)$. Then $m=\frac{-2.3 - 1.7}{-5.6-(-2.4)}=\frac{-4}{-3.2}=\frac{5}{4}=1.25$.

Step10: Solve for exercise 9

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

Step11: Solve for exercise 10

For points $(\frac{1}{2},\frac{3}{5})$ and $(-\frac{3}{4},\frac{7}{5})$, let $(x_1,y_1)=(\frac{1}{2},\frac{3}{5})$ and $(x_2,y_2)=(-\frac{3}{4},\frac{7}{5})$. Then $m=\frac{\frac{7}{5}-\frac{3}{5}}{-\frac{3}{4}-\frac{1}{2}}=\frac{\frac{4}{5}}{-\frac{5}{4}}=-\frac{16}{25}$.

Answer:

1. $m = 3$
2. $m = 4$
3. $m=-2$
4. $m = 2$
5. $m=-1.9$
6. $m=-\frac{9}{13}$
7. $m=\frac{69}{29}$
8. $m = 1.25$
9. $m=-3$
10. $m=-\frac{16}{25}$