Question

FAN and MEDAL.
Choose the point-slope form of the equation below that represents the line that passes through the point (6, -3) and has a slope of one half.

y - 6 = one half(x + 3)

y = one halfx - 6

y + 3 = one half(x - 6)

x - 2y = 12

Answer 1

@Compassionate

@DontLikeMathButOhWell

@EmeraldSunset

@Luigi0210

@taffalousswagg_bru
somebody help please.,

Answer 2

\(\bf \begin{array}{lllll}
&x_1&y_1\\
&({\color{red}{ 6}}\quad ,&{\color{blue}{ -3}})\quad
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{1}{2}\\ \quad \\

y-{\color{blue}{ y_1}}={\color{green} m}(x-{\color{red}{ x_1}})\Leftarrow\textit{point-slope form, solve for "y"}\)

Answer 3

well, based on the choices,. you don't have to solve for "y"

Answer 4

im sorry i have like no clue how to plug that in because i dont kno what to put into the un-colored bold y and x etc

Answer 5

\( \bf \begin{array}{lllll}
&\huge x_1&\huge y_1\\
&({\color{red}{ 6}}\quad ,&{\color{blue}{ -3}})\quad
\end{array}
\\\quad \\
slope = {\color{green}{ m}}= \cfrac{1}{2}\\ \quad \\
y-{\huge \color{blue}{ y_1}}={\color{green} m}(x-{\huge\color{red}{ x_1}})\Leftarrow\textit{point-slope form, solve for "y"}\)

Answer 6

see them now?

Answer 7

\( \bf \begin{array}{lllll}
&\huge x_1&\huge y_1\\
&({\color{red}{ 6}}\quad ,&{\color{blue}{ -3}})\quad
\end{array}
\\\quad \\
slope = {\color{green}{ m}}= {\huge \cfrac{1}{2}}\\ \quad \\
y-{\huge \color{blue}{ y_1}}={\huge \color{green} m}(x-{\huge\color{red}{ x_1}})\Leftarrow\textit{point-slope form, solve for "y"}\)

Answer 8

y-(-3)=m(x-6)

if this this is right then what is next?

Answer 9

well, we know m = 1/2... so \(\bf \begin{array}{lllll}
&x_1&y_1\\
&({\color{red}{ 6}}\quad ,&{\color{blue}{ -3}})\quad
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{1}{2}\\ \quad \\

y-{\color{blue}{ -3}}={\color{green} {\cfrac{1}{2}}}(x-{\color{red}{ 6}})\implies y+3=\cfrac{1}{2}x-\cfrac{6}{2}\implies y+3=\cfrac{1}{2}x-3\)

Answer 10

i get sort of why the signs changed the y+ and the -3 going to just 3, just no clue when to switch them, i get that m is 1/2 and now im lost at how the 2 got under then 6 when its 1/2x -6/2

Answer 11

but what would b the next step

Answer 12

\(\bf \cfrac{1}{2}(x-6)\implies \left(\cfrac{1}{2}\cdot x\right)+\left(\cfrac{1}{2}\cdot 6\right)\implies \cfrac{1}{2}x+\left(\cfrac{6}{2}\right)\implies \cfrac{1}{2}x+3\)

anyhow, your choices do not distribute.... so, you won't have to anyway

Answer 13

i dont think i will ever learn this , but i appreciate you helping and trying to teach me it

Answer 14

http://www.mathwarehouse.com/dictionary/D-words/distributive-property-definition-and-examples.php

Answer 15

y+3=1/2x-3, then what

Answer 16

well, check your choices, do any of them match anything you have?

Answer 17

y - 6 = one half(x + 3)

y = one halfx - 6

y + 3 = one half(x - 6)

x - 2y = 12

nope.

Answer 18

y - 6 = one half(x + 3)

y = one halfx - 6

\({\color{blue}{ \textit{y + 3 = one half(x - 6)}}}\)

x - 2y = 12

look closely

Answer 19

\(\bf \begin{array}{lllll}
&x_1&y_1\\
&({\color{red}{ 6}}\quad ,&{\color{blue}{ -3}})\quad
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{1}{2}\\ \quad \\

\bbox[border: 1px solid black]{y-{\color{blue}{ -3}}={\color{green} {\cfrac{1}{2}}}(x-{\color{red}{ 6}})}\implies y+3=\cfrac{1}{2}x-\cfrac{6}{2}\implies y+3=\cfrac{1}{2}x-3\)

Answer 20

THANK YOU, soooo much.

Answer 21

yw

Answer 22

If you wouldnt mind helping me with one last question before I get off, its okay of you dont want to,.

Answer 23

Choose the point-slope form of the equation below that represents the line that passes through the points (-3, 2) and (2, 1).

y + 3 = -5(x - 2)

y - 2 = -5(x + 3)

y + 3 = -one fifth(x - 2)

y - 2 = -one fifth(x + 3)

Answer 24

you'd need to use the distributive property for that

Answer 25

okay.

Answer 26

so, if you haven't covered it, you may want to first

Answer 27

y2-y1 , x2-x1?

Answer 28

if you have .. say 3(2+6) <-- what would that equal to?

Answer 29

24?

Answer 30

yeap... see you DISTRIBUTE the "3"

you'd do the same with the slope

so \(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ -3}}\quad ,&{\color{blue}{ 2}})\quad &({\color{red}{ 2}}\quad ,&{\color{blue}{ 1}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}
\\ \quad \\

y-{\color{blue}{ y_1}}={\color{green} m}(x-{\color{red}{ x_1}})\qquad \textit{ and you plug in your values here}\)

Answer 31

so you'd find the slope first, then grab the \(y_1\ and \ x_1\) and plug them in, along with the slope

Answer 32

y-2=m(x - -6) ?

Answer 33

well, yes, but your \(x_1\ and \ y_1\) must be from either of the GIVEN points

Answer 34

y-2=m(x - -6)
^ ^ implies a point of (-6, 2), which is not a given point

Answer 35

what would the correct way look ?

Answer 36

well,, what did you get for the slope?

Answer 37

i dont kno how to get the slope if its not given in the problem

Answer 38

unless one of the "y's" are slope i think idk

Answer 39

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ -3}}\quad ,&{\color{blue}{ 2}})\quad &({\color{red}{ 2}}\quad ,&{\color{blue}{ 1}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}
\\ \quad \\

y-{\color{blue}{ y_1}}={\color{green} m}(x-{\color{red}{ x_1}})\qquad \textit{ and you plug in your values here}\)

Answer 40

1/-5 ?

Answer 41

well... -1/5

Answer 42

so now pick a point to use, either one, which one would you like?

Answer 43

the -1 ?

Answer 44

what do you mean?

Answer 45

i thought u saud pick one of the numbers from -1/5 lol my bad as you can see idk what to do at all

Answer 46

i found the answer thank you for everything