Question

Find the equation of the straight line which has y-intercept equal to $\frac{4}{3}$ and is perpendicular to $3x-4y+1=0.$

Answer 1

Any line having y-itnercept equal to $\frac{4}{3}$

Passes through the point $\begin{array}{c}(0,\frac{4}{3})\\ (x_1,y_1)\end{array}$

Slope of line $3x-4y+11=0$

$y=\frac{3}{4}x+\frac{11}{4}$

$\Rightarrow m=\frac{3}{4}$

The required line is perpendicular to the given line, therefore its slope is $\frac{-4}{3}$

$\Rightarrow$ Equation of required line is

$y-y_1=m^{\prime }(x-x_1)$

$y-\frac{4}{3}=\frac{-4}{3}(x-0)$

$4x+3y-4=0$