Question

find the equation of the line perpendicular to the given line and passing through the given p
$2x + 3y = 6$, $(3, -2)$
answer

Explanation:

Step1: Find slope of given line

Rewrite $2x+3y=6$ to slope-intercept form $y=mx+b$:
$3y = -2x + 6$
$y = -\frac{2}{3}x + 2$
Slope of given line $m_1 = -\frac{2}{3}$

Step2: Find perpendicular slope

Perpendicular slope $m_2$ satisfies $m_1 \times m_2 = -1$:
$m_2 = \frac{-1}{-\frac{2}{3}} = \frac{3}{2}$

Step3: Use point-slope form

Point-slope formula: $y - y_1 = m_2(x - x_1)$ with $(x_1,y_1)=(3,-2)$:
$y - (-2) = \frac{3}{2}(x - 3)$

Step4: Simplify to standard form

$y + 2 = \frac{3}{2}x - \frac{9}{2}$
Multiply all terms by 2 to eliminate fractions:
$2y + 4 = 3x - 9$
Rearrange to $Ax+By=C$:
$3x - 2y = 13$

Answer:

$3x - 2y = 13$