Question

The equation of a line through (3, –4) and perpendicular to the line 3x + 4y = 5 is

• A. 4x + 3y = 24
• B. y – 4 = (x + 3)
• C. 3y – 4x = 24
• D. $y+4=\frac{4}{3}(x-3)$

Answer 1

The correct option is D $y+4=\frac{4}{3}(x-3)$

Slope of the given line is $\frac{-3}{4}$. So the slope of the required line will be $\frac{4}{3}$
The required equation, which passes through (3, –4) and its gradient is $\frac{4}{3}$, is $(y+4)=\frac{4}{3}(x-3)$.