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Find the equation of the line passing through the point (– 1, 3) and perpendicular to the line 3x – 4y – 16 = 0.

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The given equation is

 3x – 4y – 16 = 0 

⇒ y = \(\frac{3}{4}\) x – 4 

∴ Slope of the line, m1 = \(\frac{3}{4}\) 

∴ Slope of the perpendicular line, m2 = – \(\frac{4}{3}\) . 

Since the line passes through (– 1, 3), so the equation of the line is 

y – y0 = m(x – x0

⇒ y – 3 = m2 (x+ 1) 

⇒ y – 3 = –\(\frac{4}{3}\)(x + 1) 

⇒ 3y – 9 = – 4x – 4 

⇒ 4x + 3y – 5 = 0

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