To solve the problem of finding the equation of the line passing through ((2,3)) and perpendicular to (2x - 3y + 6 = 0), follow these steps:

Step 1: Find the slope of the given line

Rewrite the given line in slope-intercept form (y = mx + b):
[2x - 3y + 6 = 0]
[ -3y = -2x -6]
[y = \frac{2}{3}x + 2]
Slope of the given line (m_1 = \frac{2}{3}).

Step 2: Find the slope of the perpendicular line

The slope of a line perpendicular to (m_1) is the negative reciprocal:
[m_2 = -\frac{3}{2}]

Step 3: Use point-slope form to find the equation

Point-slope formula: (y - y_1 = m(x - x_1)) where ((x_1,y_1)=(2,3)) and (m=-\frac{3}{2}):
[y - 3 = -\frac{3}{2}(x - 2)]

Multiply both sides by 2 to eliminate the fraction:
[2(y - 3) = -3(x - 2)]
[2y - 6 = -3x + 6]
[3x + 2y - 12 = 0]

Answer: (\boxed{3x + 2y - 12 = 0}) or (\boxed{3x + 2y = 12}) (either form is acceptable).

(\boxed{3x + 2y = 12}) is the simplified form.

(\boxed{3x + 2y = 12})

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