Question

The equation of the line whose $y-$intercept is $3/4$ and which is perpendicular to $3x+4y+11=0$

• A. $4x-3y+4=0$
• B. $6x+4y+3=0$
• C. $6x-4y-3=0$
• D. $6x+4y-3=0$

Answer 1

The correct option is A $4x-3y+4=0$

Let the equation of the line be $\frac{x}{a}+\frac{y}{\frac{4}{3}}=1$......(1) [ Since $y$ intercept is $\frac{4}{3}$ (given)]

Now equation (1) can be written as $y=-\frac{4}{3a}x+\frac{4}{3}$.

So the gradient of the line is $\frac{-4}{3a}$.

Given the equation of the line is $3x+4y+11=0$ this line can be written as $y=-\frac{3}{4}x-\frac{11}{4}$......(2).

The gradient of the line (2) is $-\frac{3}{4}$.

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

$-\frac{4}{3a}=\frac{4}{3}$

or, $a=-1$.

The equation of the line is $\frac{x}{-1}+\frac{3x}{4}=1$ or, $4x-3y+4=0$.