Question

12.) Write an equation of the line in slope-intercept form with slope 2, passing through (4, -1). (Be sure to show all your work.)



13.) Write an equation of the line in point-slope form passing through (1, 5) and (3, 11). (Be sure to show all your work.)



14.) What is an equation of the line perpendicular to y = -4x + 1 that contains (2, -3)? (Be sure to show all your work.)



15.) Write an equation of a vertical line passing through (-1, 3). (Be sure to show all your work.)

Answer 1

12 \(\bf \begin{array}{lllll}
&x_1&y_1\\
&(4\quad ,&-1)
\end{array}
\\\quad \\

slope = m= 2
\\ \quad \\

y-y_1=m(x-x_1)\quad \textit{plug in your values and solve for "y"}\)

Answer 2

13 \(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&(1\quad ,&5)\quad &(3\quad ,&11)
\end{array}
\\\quad \\

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

y-y_1=m(x-x_1)
\quad \textit{plug in your values and solve for "y"}\)

Answer 3

i dont know how or what u meen

Answer 4

pleease this is a matter of passing or failing

Answer 5

i have 13 ut i need th rest

Answer 6

and i know 12 so all i need is 14 and 15

Answer 7

"14.) What is an equation of the line perpendicular to y = -4x + 1 that contains (2, -3)? (Be sure to show all your work.)

first find the slope of the perpendicular line, can you do this using the equation
\[m_am_b=-1\]

Answer 8

i have no clue how to do that

Answer 9

y = -4x + 1
whats the slope in this?

Answer 10

idk

Answer 11

"pleease this is a matter of passing or failing" <--- the idea behind exercises is that you understand the material, the grading system is there to assess your understanding

if you just want the answer or someone to do it for you, well... the diverges from the meaning of "help" for many folks here

Answer 12

slope is the value of the coefficient that is infront of the 'x' in the equation
y = -4x + 1

so what is the number that is infront of the x?

Answer 13

if your calling me a lier your rong im passing one class and its not this one and if i just wanted the answere i whouldent be here so thats realy disrepectfull to do thanks bey way thow

Answer 14

-4

Answer 15

so -4 is the slope of the line, y = -4x + 1

now let m_a = -4

plug it into,
m_a * m_b =-1
and solve for m_b

Answer 16

and how do i do that?

Answer 17

so the slope of the equation y = -4x + 1 is -4
a perpendicular line to it will have a NEGATIVE RECIPROCAL slope

NEGATIVE of that
+4
RECIPROCAL of that
1/4

so now you have a line whose slope is 1/4 and passes through the points of (2, -3)
\(\bf \begin{array}{lllll}
&x_1&y_1\\
&(2\quad ,&3)
\end{array}
\\\quad \\

slope = m= \cfrac{1}{4}
\\ \quad \\

y-y_1=m(x-x_1)
\quad \textit{plug in your values and solve for "y"}\)

Answer 18

ok now 15

Answer 19

yea

Answer 20

this one will have RUN of ... the same... since x = -1 no matter where you go

Answer 21

so lemme use another point... say (-1, 2) to go through the slope process... so we get \(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&(-1\quad ,&3)\quad &(-1\quad ,&2)
\end{array}
\\\quad \\

slope = m= m= \cfrac{rise}{run} \implies \cfrac{(2)-(3)}{(-1)-(-1)}\implies \cfrac{-1}{0}\implies undefined\)

Answer 22

anyhow to make it short... not matter what point you pick, RUN will allways be undefined, and thus \(\begin{array}{llll}
y=\pm 5&x =-1\\
y=\pm 5000&x =-1\\
y=\pm 5,000,000&x =-1\\
y=\pm \infty&x =-1

\end{array}\)

so in short