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 $O A B$ is minimum ( $O$ is origin), then which of the following statement(s) is(are) correct?

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

Answer 1

Answer: (A), (D)

Let the line be
$\frac{ x }{ a }+\frac{ y }{ b }=1, a , b >0$
where $\frac{1}{ a }+\frac{4}{ b }=1 \Rightarrow \frac{4}{ b }=1-\frac{1}{ a } \Rightarrow \frac{4}{ b }=\frac{ a -1}{ a }$
$b=\frac{4 a}{a-1}$
Now, $2 A = ab = a \left(\frac{4 a }{ a -1}\right)=\frac{4\left( a ^2-1+1\right)}{ a -1}=4\left[( a +1)+\frac{1}{ a -1}\right]$

$2 \frac{ dA }{ da }=4\left[1-\frac{1}{( a -1)^2}\right]=0 $
$a -1=1 \text { or }-1 $
$a =2 ; b =8, m =-4$
$A ]_{\min }=\frac{16}{2}=8$
Equation is $(y-4)=-4(x-1) \Rightarrow y-4=-4 x+4 \Rightarrow 4x + y = 8$
Also, $r=\frac{\Delta}{ s }=\frac{\frac{1}{2} \times 2 \times 8}{\frac{2+8+\sqrt{68}}{2}}=\frac{8}{5+\sqrt{17}}=5-\sqrt{17}$
Now verify alternatives.