Question

question
find the equation of a line perpendicular to $3x + y = 3$ that passes through the point $(6, 5)$.
answer
$\bigcirc \\ x + 3y = 21$
$\bigcirc \\ y = \frac{1}{3}x + 3$
$\bigcirc \\ y = -\frac{1}{3}x + 3$
$\bigcirc \\ 3x + y = 23$
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Explanation:

Step1: Find slope of given line

Rewrite $3x + y = 3$ to slope-intercept form:
$y = -3x + 3$, so slope $m_1 = -3$

Step2: Find perpendicular slope

Perpendicular slope $m_2 = -\frac{1}{m_1} = \frac{1}{3}$

Step3: Use point-slope form

Point-slope formula: $y - y_1 = m_2(x - x_1)$
Substitute $(x_1,y_1)=(6,5)$ and $m_2=\frac{1}{3}$:
$y - 5 = \frac{1}{3}(x - 6)$

Step4: Simplify to slope-intercept form

$y - 5 = \frac{1}{3}x - 2$
$y = \frac{1}{3}x + 3$

Step5: Verify other form (optional)

Rewrite to standard form: $x - 3y = -9$, which does not match $x+3y=21$. Confirm the slope-intercept form is correct.

Answer:

$y = \frac{1}{3}x + 3$