Question

If a straight line $L$ with negative slope passes through $P(1,4)$ and meets the coordinate axes at $A$ and $B$ such that area of triangle $OAB$ is minimum $(O$ is origin$),$ then which of the following statements is (are) CORRECT ?

• A. $y-$intercept of line $L$ is $2.$
• B. Distance of line $L$ from $(\sqrt{4},\sqrt{17})$ is $1$ unit
• C. Area of triangle $OAB$ is $16$ sq. units
• D. Radius of the circle inscribed in triangle $OAB$ is $5-\sqrt{17}$ units

Answer 1

Answer: (B), (D)

Radius of the circle inscribed in triangle $OAB$ is $5-\sqrt{17}$ units

Let the line equation be
$\frac{x}{a}+\frac{y}{b}=1,a,b>0\cdots \left(i\right)$
Since line passes through $P(1,4),$
$\Rightarrow \frac{1}{a}+\frac{4}{b}=1$
$\Rightarrow b=\frac{4a}{a-1}$

Let Area of $△OAB=\Delta$
Now, $2\Delta =ab=a\left(\frac{4a}{a-1}\right)=\frac{4a^2}{a-1}$
$2\frac{d\Delta }{da}=4\left[\frac{(a-1)2a-a^2\cdot 1}{{(a-1)}^2}\right]=0\Rightarrow a=0,2$
and $b=0,8$ respectively
$\because a,b>0\Rightarrow a=2,b=8,m=-4$
So, ${\Delta }_{min}=\frac{16}{2}=8$ sq. units

From $\left(i\right),$
line equation is $\frac{x}{2}+\frac{y}{8}=1$
$\Rightarrow 4x+y-8=0$
So, distance of line $L$ from $(\sqrt{4},\sqrt{17})$ is $\left|\frac{4\sqrt{4}+\sqrt{17}-8}{\sqrt{16+1}}\right|=1$ unit

$\Delta =8,s=\frac{2+8+\sqrt{68}}{2}=5+\sqrt{17}$
$\therefore r=\frac{\Delta }{s}=\frac{8}{5+\sqrt{17}}=5-\sqrt{17}$ units