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Find the equation of a line passing through (3, -2) and perpendicular to the line x – 3y + 5 = 0. |
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Answer» Given: equation is perpendicular to x – 3y + 5 = 0 and passes through (3,-2) To find: Equation of required line. Explanation: The equation of the line perpendicular to x − 3y + 5 = 0 is 3x + y + λ = 0, Where λ is a constant. It passes through (3, − 2). 9 – 2 + λ = 0 ⇒ λ = - 7 Substituting λ = − 7 in 3x + y + λ = 0, Hence, we get 3x + y – 7 = 0, which is the required line. |
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