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.

Answer 1

The line perpendicular to 3x − 4y + 11 = 0 is \[4x + 3y + \lambda = 0\]

It is given that the line

\[4x + 3y + \lambda = 0\]  has y-intercept equal to \[\frac{4}{3}\] 

This means that the line passes through \[\left( 0, \frac{4}{3} \right)\]

\[\therefore 0 + 4 + \lambda = 0\]

\[ \Rightarrow \lambda = - 4\]

Substituting the value of

\[\lambda\] , we get 

\[4x + 3y - 4 = 0\],which is equation of the required line.