Question

A straight line $L$ with negative slope passes through the point $(1,1)$ and cuts the positive coordinate axes at the points $A$ and $B$. If $O$ is the origin, then the minimum value of $OA+OB$ as $L$ varies, is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

Equation of line having slope ${}^{\prime }{m}^{\prime }$ passes through the point $(1,1)$ is
$y-1=m(x-1)$ ...(i)
So, $A\left(\frac{m-1}{m},0\right)$ and $B(0,1-m)$
Now, $OA+OB=\left(1-\frac{1}{m}\right)+(1-m)=2-\left(m+\frac{1}{m}\right)$
$\because m$ is negative, so minimum value of
$-\left(m+\frac{1}{m}\right)=2$
So, minimum value of $OA+OB=2+2=4$