Question

PLEaseEee help!


Select the equations that are parallel and perpendicular to y = x + 5 and that pass through the point (-2, -1).

parallel: y = -x - 1
perpendicular: y = x + 2

parallel: y = x - 1
perpendicular: y = -x + 1

parallel: y = x + 1
perpendicular: y = -x - 3

parallel: y = 2x - 2
perpendicular: y = -2x - 1

Answer 1

what's the slope of say " y = x + 5 " anyway?

Answer 2

Um, im not exactly sure..

Answer 3

-x+y-5=0?

Answer 4

\(\large \bf y = x + 5\implies y = {\color{blue}{ 1}}x + 5\) see the slope now?

Answer 5

im gonna say its 1 haha

Answer 6

hehe right or 1/1 = 1 rise/run meaning rise =1 run =1

so any parallel line to that one.. will also have the same slope of 1
so to find an equation of a parallel line to it, you're looking for an equation whose slope is 1 and passes through (-2, -1)
\(\bf \begin{array}{lllll}
&x_1&y_1\\
&({\color{red}{ -2}}\quad ,&{\color{blue}{ -1}})\quad
\end{array}
\\\quad \\

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

y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\
\qquad \uparrow\\
\textit{point-slope form}\)
------------------------------------------------------

now...a line that's perpendicular to y = x + 5, will have the NEGATIVE RECIPROCAL slope of it
so y = x + 5 has a slope of 1 or 1/1 then \(\bf \cfrac{{\color{blue}{ 1}}}{1}\qquad negative\implies -\cfrac{{\color{blue}{ 1}}}{1}\qquad reciprocal\implies -\cfrac{1}{{\color{blue}{ 1}}}\implies -1\)

so you're looking for the equation of a line that has a slope of -1 and passes through (-2, -1)

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

slope = {\color{green}{ m}}= -1
\\ \quad \\

y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\
\qquad \uparrow\\
\textit{point-slope form}\)

Answer 7

Woah you really know you're stuff :O sadly i do not and im still confused LOL

Answer 8

hold on let me try and solve for y

Answer 9

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

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

y-{\color{blue}{ (-1)}}={\color{green}{ 1}}(x-{\color{red}{ (-2)}})\implies y+1=1(x+2)\implies y+1=x+2\\
\qquad \uparrow\\
\textit{point-slope form}
\\ \quad \\
y=?\)

Answer 10

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

slope = {\color{green}{ m}}= -1
\\ \quad \\

y-{\color{blue}{ (-1)}}={\color{green}{ -1}}(x-{\color{red}{ (-2)}})\implies y+1=-1(x+2)\implies y+1=-x-2\\
\qquad \uparrow\\
\textit{point-slope form}
\\ \quad \\
y=?\impliedby perpendicular\)

Answer 11

oh my um um i feel so silly that im still confused

Answer 12

well... once you plug in the values... .is just a matter of linear simplification, which the exercise assumes you know already

Answer 13

for the first would it be y=x+1

Answer 14

yeap thus " y=x+1 " is parallel to y = x + 5

Answer 15

Oh yay! thank you so much! you think it would be bothersome to help me with one more?

Answer 16

you can always post anew... thus if I dunno, someone else may, and we can revise each other