Question

What is the equation of the line that passes through (12, 4) and is perpendicular to the graph of y = -3 over 4x - 2?

Answer 1

Giving medals also, and fan!

Answer 2

what is the slope of \(\bf y = -\cfrac{3}{4}x-2\quad ?\)

Answer 3

m=-3/4

Answer 4

right, so the slope of a line PERPENDICULAR to that one will be the NEGATIVE RECIPROCAL of that, what does that mean?

well

NEGATIVE of -3/4 ?
3/4

RECIPROCAL of that?
4/3

so \((\bf \begin{array}{lllll}
&x_1&y_1\\
&(16\quad ,&4)
\end{array}
\\\quad \\

slope = m= \cfrac{4}{3}
\\ \quad \\

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

Answer 5

16? darn... typo

Answer 6

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

slope = m= \cfrac{4}{3}
\\ \quad \\

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

Answer 7

y = -3/4x + 13?

Answer 8

wait
wait
wait

Answer 9

y = 4 over 3x - 12?

Answer 10

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

slope = m= \cfrac{4}{3}
\\ \quad \\

y-y_1=m(x-x_1)\implies y-12=\cfrac{4}{3}(x-12)\\ \quad \\
y-12=\cfrac{4}{3}x-\cfrac{4\cdot 12}{3}\implies y-12=\cfrac{4}{3}x-\cfrac{48}{3}\implies y-12=\cfrac{4}{3}x-16\\ \quad \\
y=\cfrac{4}{3}x-16+12
\)

Answer 11

\(\bf y-y_1=m(x-x_1)\implies y-12=\cfrac{4}{3}(x-12)\\ \quad \\
y-12=\cfrac{4}{3}x-\cfrac{4\cdot 12}{3}\implies y-12=\cfrac{4}{3}x-\cfrac{48}{3}
\\ \quad \\y-12=\cfrac{4}{3}x-16\\ \quad \\
y=\cfrac{4}{3}x-16+12
\) anyhow, since it got a bit truncated

Answer 12

So I was correct?

Answer 13

well... .-16 + 12 \(\ne -12\)

Answer 14

-12 is part of my answer :D