Question

The equation of line through $(3,4)$ which cuts from the first quadrant a triangle of minimum area is

• A. $4x+3y=12$
• B. $3x+4y=12$
• C. $2x+3y=13$
• D. $3x+2y=24$

Answer 1

The correct option is A $4x+3y=12$

Let the x-intercept of line be 'a' and y-intercept of the line be 'b'

Equation of line : $\frac{x}{a}+\frac{y}{b}=1$

It passes through $(3,4)$

$\Rightarrow \frac{3}{a}+\frac{4}{b}=1$

Area with the axis $=\frac{1}{2}ab=\frac{1}{2}a\left(\frac{4}{\frac{3}{a}-1}\right)=\frac{2a^2}{3-a}$

Given area is minimum $\Rightarrow \frac{d}{da}\left(area\right)=0$

$\Rightarrow (3-a)\left(4a\right)-2a^2(-1)=0$

$\Rightarrow a=6$

Substituting above we get ; $b=8$

Equation of line : $\frac{x}{6}+\frac{y}{8}=1$

$\Rightarrow$ Equation of line : $4x+3y=12$