Question

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

• A. $8$
• B. $16$
• C. $24$
• D. $32$

Answer 1

The correct option is C $24$

We have,

Given points are $(3,4)$

Meets the axes $OX$ and $OY$at $A$ and $B$ respectively.

We know that

Equation of line is $y=mx+c$

Its passes through the points $(3,4)$

Then,

$4=3m+c......\left(1\right)$

Now area of

$\Delta OAB=\frac{1}{2}\times OA\times OB$

$=\frac{1}{2}\times \left(\frac{-c}{m}\right)\times c$

From equation (1) and we get,

$4=3m+c$

$3m=4-c$

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

$Areaof\Delta OAB=\frac{1}{2}\times \frac{-c^2}{\left(\frac{4-c}{3}\right)}$

$A=\frac{-3c^2}{8-2c}$

For minimum area

Differentiate this w.r.t c and we get,

$\frac{dA}{dc}=\frac{(8-2c)\frac{d}{dx}(-3c^2)-(-3c^2)\frac{d}{dx}(8-2c)}{{(8-2c)}^2}$

$=\frac{(8-2c)(-6c)-6c^2}{{(8-2c)}^2}$

$=\frac{-48c+12c^2-6c^2}{{(8-2c)}^2}$

$=\frac{6c^2-48c}{{(8-2c)}^2}$

For maximum and minimum,

$\frac{dA}{dc}=0$

$\Rightarrow \frac{6c^2-48c}{{(8-2c)}^2}=0$

$\Rightarrow 6c^2-48c=0$

$\Rightarrow c=0,c=8$

Put $c=8$ in the required area

$Area=\frac{-3\times {\left(8\right)}^2}{8-2\times 8}$

$=\frac{-3\times 8\times 8}{-8}$

$=24sq.unit.$

Hence, this is the answer.