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
1) 1
2) 2
3) 3
4) 4​

Answer 1

Let the angle of inclination of the line be $\theta$ with positive x axis, which is an obtuse angle since the line has negative slope.

Let the points of intercepts A and B be at positive x and y axes respectively, so that their coordinates are taken as (a, 0) and (0, b) respectively.

Therefore,

• $OA=a$

• $OB=b$

Combining the points (a, 0) and (1, 1), the slope can be written as,

$\longrightarrow\tan\theta=\dfrac{1-0}{1-a}$

$\longrightarrow\tan\theta=\dfrac{1}{1-a}$

Taking the reciprocal we get,

$\longrightarrow\cot\theta=1-a$

$\longrightarrow a=1-\cot\theta$

Combining the points (0, b) and (1, 1), the slope can be written as,

$\longrightarrow\tan\theta=\dfrac{1-b}{1-0}$

$\longrightarrow\tan\theta=1-b$

$\longrightarrow b=1-\tan\theta$

Then, value of OA + OB will be,

$\longrightarrow a+b=(1-\cot\theta)+(1-\tan\theta)$

$\longrightarrow a+b=2-\tan\theta-\cot\theta\quad\quad\dots(1)$

To find minimum value of $a+b,$ we can equate derivative of $a+b$ wrt $\theta$ as zero.

$\longrightarrow\dfrac{d}{d\theta}(a+b)=0$

$\longrightarrow\dfrac{d}{d\theta}(2-\tan\theta-\cot\theta)=0$

$\longrightarrow0-\sec^2\theta-(-\csc^2\theta)=0$

$\longrightarrow\csc^2\theta-\sec^2\theta=0$

$\longrightarrow\csc^2\theta=\sec^2\theta$

$\longrightarrow\dfrac{\sec^2\theta}{\csc^2\theta}=1$

$\longrightarrow\tan^2\theta=1$

Since $\theta$ is an obtuse angle,

$\longrightarrow\tan\theta=-1$

$\Longrightarrow\theta=135^o$

Then (1) becomes,

$\longrightarrow a+b=2-\tan135^o-\cot135^o$

$\longrightarrow a+b=2-(-1)-(-1)$

$\longrightarrow\underline{\underline{a+b=4}}$

Hence (4) is the answer.