Find the equation of the straight line which has y-intercept equal to 4/3 and is perpendicular to `3x-4y+11=0.`

Explanation

equation (1) can be written as `y=4(−3a)x+4/3​.`

So the gradient of the line is `(-4)/(3a)​`.

Given the equation of the line is` 3x+4y+11=0`
this line can be written as `y=−4/3​x−4/(11)​......(2).`

The gradient of the line (2) is `−4/3​`.

Since line (1) is given to be perpendicular to (1) then we get,

`4/(−3a)​=4/3​`

or, `a=−1`.

The equation of the line `is −x​+3/4x​=1 or, 4x−3y+4=0.`