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Question
Find the equation of the straight line perpendicular to 2x − 3y = 5 and cutting off an intercept 1 on the positive direction of the x-axis.
Solution
The line perpendicular to 2x − 3y = 5 is \[3x + 2y + \lambda = 0\]
It is given that the line \[3x + 2y + \lambda = 0\] cuts off an intercept of 1 on the positive direction of the x-axis.
This means that the line \[3x + 2y + \lambda = 0\] passes through the point (1, 0).
\[\therefore 3 + 0 + \lambda = 0\]
\[ \Rightarrow \lambda = - 3\]
Substituting the value of \[\lambda\],we get
\[3x + 2y - 3 = 0\] which is equation of the required line.
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