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Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. |
Answer» 3\(x\) – 2y + 5 = 0 ⇒ –2y = –3\(x\) – 5 ⇒ y = \(\frac{3}{2}\)\(x\) + \(\frac{5}{2}\) On comparing with y = m\(x\) + c, we see that slope of given line = \(\frac{3}{2}\) As the required line is perpendicular to the given line, Slope of required line = \(\frac{-3}{2}\) ∴ Equation of required line: (y – 5) = \(\frac{-3}{2}\) (\(x\) - 4) ⇒ 3(y – 5) = – 2\(x\) + 8 ⇒ 3y – 15 = –2\(x\) + 8 ⇒ 3y + 2\(x\) – 23 = 0 |
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