Question

The equation of line CD is y = 3x − 3. Write an equation of a line perpendicular to line CD in slope-intercept form that contains point (3, 1).

y = 3x + 0

y = −3x − 8

y = negative 1 over 3x + 2

y = − 1 over 3x + 0
@tester97

Answer 1

hmmm what's the slope of => y = 3x − 3. ?

Answer 2

3x?? @jdoe0001

Answer 3

\(\bf \begin{array}{llll}
y = &{\color{brown}{ 3}}x - 3\\
&\uparrow \\
&{\color{brown}{ slope}}
\end{array}\)

so the slope of a line perpendicular to this one, will have a slope that is, NEGATIVE RECIPROCAL to this slope
so this one is 3 so \(\bf 3\qquad negative\to -3 \qquad reciprocal\to \cfrac{1}{-3}\)

so what you're really asked is, to find the equation of a line whose slope is -1/3 and passes through (3,1) thus \(\bf \begin{array}{lllll}
&x_1&y_1\\
&({\color{red}{ 3}}\quad ,&{\color{blue}{ 1}})\quad
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= -\frac{1}{3}
\\ \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 4

solving for "y" will more or less give you the -> y = mx+b <-- form, so-called the slope-intercept form

Answer 5

i still dont get it :( @jdoe0001

Answer 6

ok... what part?

Answer 7

the plug in numbers.. @jdoe0001

Answer 8

well, notice the values, they're pretty much given

Answer 9

sooo it would be c?? @jdoe0001