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Question
Write the equation of the lines through the point (1, −1) perpendicular to 3x + 4y = 6
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Solution
Any line perpendicular to 3x + 4y – 6 = 0 will be of the form 4x – 3y + k = 0.
It passes through (1, –1)
⇒ 4 + 3 + k = 0
⇒ k = – 7.
So the required line is 4x – 3y – 7 = 0
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