Question

A straight line segment through the point $(3,4)$ in the first quadrant meets the coordinate axes in A and B. The minimum area of $\Delta AOB$ is

• A. $42$
• B. $64$
• C. $48$
• D. $24$

Answer 1

Answer: (D)

$24$

Equation of straight line passing through $(3,4)$ and having slope $m$ is given
$y-4=m(x-3)$
Since, this line intersects coordinate axes at A and B.
So, coordinates of A are $(\frac{3m-4}{m},0)$
Coordinates of B are $(0,-3m+4)$
Area of $△AOB=-\frac{1}{2}\frac{(3m-4{)}^2}{m}$
$A=f\left(m\right)=-\frac{1}{2}\left(\frac{9m^2-24m+16}{m}\right)$
For maxima or minima,
$f^{\prime }\left(m\right)=0$
$\Rightarrow \frac{9m^2-16}{m^2}=0$
$\Rightarrow m=\pm \frac{4}{3}$
$f^{\prime \prime }\left(x\right)=-\frac{32}{m^3}$
$f^{\prime \prime }\left(x\right)>0$ at $m=-\frac{4}{3}$
Hence, area is minimum at $m=\frac{4}{3}$
So, Area $=24$ sq.units