Question

State the following statement is True or False

$A$ line passing through $(3,4)$ meets the axes $OX$ and $OY$ at $A$ and $B$ respectively. The minimum area of the triangle $OAB$ in square units is $34$

• A. True
• B. False

Answer 1

The correct option is B False

Equation of line AB,

$y-4=m(x-3)$ for $m<0$

$y=mx-3m+4$

So, $x=0,y=4-3m$

$y=0,x=\frac{3m-4}{m}$

$\therefore$ area of $△OAB=\frac{1}{2}\times \left(\frac{3m-4}{m}\right)\times (4-3m)$

$=\frac{12m-9m^2-16+12m}{2m}$

$=\frac{24m-9m^2-16}{2m}$

$=\frac{24}{2}-(\frac{9}{2}m+8\frac{1}{m})$

For minimum area, $\frac{9m}{2}+\frac{8}{m}$ is maximum,

$\therefore f\left(m\right)=\frac{9m}{2}+\frac{8}{m}$

$f^1\left(m\right)=\frac{9}{2}-\frac{8}{m^2}=0$

$m^2=\frac{16}{9}$

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

For minimum area, $m=\frac{-4}{3}$

Minimum area of $\Delta =12-\left(\frac{9}{2}\right(\frac{-4}{3})-6)$

$=18+6=24$