Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th
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There are two ways to prove it. 1st way: Area of triangle formed by the given points = 0 if they are collinear.∴ \(rac{1}{2}\) [\(x\)(2 – (y + 1) ) + 1((y + 1) – 1) + 0(1 – 2)] = 0⇒ \(rac{1}{2}\) [2\(x\) – \(x\)y – \(x\) + y + 0] = 0 ⇒ \(x\) + y – \(x\)y = 0 ⇒ \(x\) + y = \(x\)y ⇒ \(rac{x}{xy}\) + \(rac{y}{xy}\) = 1 ⇒ \(rac{1}{y}\) + \(rac{1}{x}\) = 1.2nd way: Slope of the lines formed by joining these points are equal. Slope of line joining (\(x\), 1), (1, 2) = \(rac{(2-1)}{1-x}\)Slope of line joining (1, 2), (0, y + 1) = \(rac{y+1-2}{0-1}\)⇒ \(rac{1}{1-x}\) = \(rac{y-1}{-1}\) ⇒ -1 = (1 – x) (y – 1) ⇒ –1 = y – \(x\)y – 1 + \(x\) ⇒ x + y = \(x\)y ⇒  \(rac{1}{x}\) + \(rac{1}{y}\) = 1.
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