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Question
Find the equations of straight lines which are perpendicular to the line 3x + 4y − 6 = 0 and are at a distance of 4 units from (2, 1)
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Solution
The equation of the given line is
3x + 4y – 6 = 0 ......(1)
The equation of any line perpendicular to line (1) is
4x – 3y + k = 0 ......(2)
Given that this line is 4 units from the point (2, 1)
∴ 4 = `+- (4(2) - 3(1) + k")/sqrt(4^2 + ( - 3)^2`
± 4 = `(8 - 3 + "k")/sqrt(16 + 9)`
± 4 = `(5 + "k")/5`
5 + k = ± 20
k = ± 20 – 5
k = 20 – 5 or k = -20
k = 15 or k = – 25
∴ The equation of the required lines are
4x – 3y + 15 = 0 and 4x – 3y – 25 = 0
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