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Question
Find the equation of the straight line which has y-intercept equal to \[\frac{4}{3}\] and is perpendicular to 3x − 4y + 11 = 0.
Solution
The line perpendicular to 3x − 4y + 11 = 0 is \[4x + 3y + \lambda = 0\]
It is given that the line
\[\therefore 0 + 4 + \lambda = 0\]
\[ \Rightarrow \lambda = - 4\]
Substituting the value of
\[\lambda\] , we get
\[4x + 3y - 4 = 0\],which is equation of the required line.
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