Question

A line passes through the points (-8, 7) and (x2, y2), and has a slope of -1/10. If point (x2, y2) is located in quadrant I, find x2.

Answer 1

4) A line passes through the points (-8,7) and (x2, y2), and has a slope of 1/14. If point (x2, y2) is located in quadrant I, find y2.

y = (1/14)x + b passes through (8,-7), so
7 = (1/14)(-8) + b --> 106/14

y = (1/14)x + 106/14

This line passes through infinitely many points in quadrant I

Answer 2

uh wats this

Answer 3

the asnwer

Answer 4

my slope is different

Answer 5

oh

Answer 6

my bad

Answer 7

and is find x2

Answer 8

its ok

Answer 9

well srry for trying to help

Answer 10

jk derp

Answer 11

so, do you know the answer

Answer 12

y = mx + b

y = (-1/10)x + b

7 = (-1/10)(-8) + b

7 = 8/10 + b

solve for b

Answer 13

so is b 6.2

Answer 14

keep it as a fraction

Answer 15

so its 6 1/5

Answer 16

x2=6 1/5

Answer 17

am i right

Answer 18

that's the value of b, not x2

Answer 19

b = 6 1/5 or b = 31/5

Answer 20

but the question says find x2

Answer 21

so your equation is

\[\Large y = -\frac{1}{10}x + \frac{31}{5}\]

Answer 22

you can then use this to find the x-intercept

Answer 23

how

Answer 24

by plugging in y = 0 and solving for x

Answer 25

once you have the x-intercept, you can determine the range where the line is in Q1

Answer 26

which will help you find x2 (there are many values to choose from for x2)

Answer 27

so if i solve this equation 0=-1/10x+ 31/5 i will find x2

Answer 28

that will give you the largest possible value you can pick for x2

Answer 29

you basically have this line

Answer 30

the point (x2, y2) will lie on the line and will lie in Q1

Answer 31

so you can see why there are infinitely many possibilities

Answer 32

how do you now its a negative line not a positive line

Answer 33

because you have a negative slope

Answer 34

-1/10 is negative

Answer 35

oh yeah you said if i plug y it will give me the largest possible value you can pick for x2 how can i make it the least possible value

Answer 36

x = 0 is the smallest possible value for x2

Answer 37

so that part is taken care of

Answer 38

ok thanks for the help i kinda know it but can i ask you question tommorrow

Answer 39

sure