y - 2 = -\frac{4}{3}(x - 1)

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• 3x + 12 = 4y \Rightarrow y = \frac{3}{4}x + 3
• So slope \( m = \frac{3}{4} \). The perpendicular slope is \( -\frac{4}{3} \).
• The line through \( (1, 2) \) with slope \( -\frac{4}{3} \) is:
• \Rightarrow y = -\frac{4}{3}x + \frac{4}{3} + 2 = -\frac{4}{3}x + \frac{10}{3}
• Now find intersection with original line \( 3x - 4y + 12 = 0 \).
• Substitute \( y = -\frac{4}{3}x + \frac{10}{3} \):

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