Question

through (2,-3) and parallel to x-5y=10

Answer 1

Luigi :3

Answer 2

Get the equation in slope intercept form, find the slope, m, then use \(\Large y-y_{1}=m(x-x_{1})\)

\[\Large \color{green}{Batman~}\]

Answer 3

so i would set it up as y+3=1/5(x-2)?

Answer 4

if you replace x =2, y =-3 into the left hand side of the expression, what do you have?

Answer 5

*cough* what's the slope of " x-5y=10 " ?

Answer 6

-1/5x is the slope

Answer 7

I know my method is weird, but it is 100% correct for this kind of problem
2-5*(-3) = 17
then x-5y = 17 is the required line.

Answer 8

so... what would it be if we solve "x-5y=10" for y?

Answer 9

y=1/5x-2

Answer 10

so our slope will be \(\bf y={\color{brown}{ \cfrac{1}{5}}}x-2\)

Answer 11

yes

Answer 12

so a parallel line will have the same slope as that one, or 1/5 so you're really being asked to find the equation of a line with slope of 1/5 and with point of (2,-3) \(\bf \begin{array}{lllll}
&x_1&y_1\\
&({\color{red}{ 2}}\quad ,&{\color{blue}{ -3}})\quad
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{1}{5}
\\ \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 13

y+3=1/5(x-2)

Answer 14

yeap

Answer 15

so when the distributive property is applied then it would be:
\[y+3=1/5x-2/5\]

Answer 16

yeap