Question

A straight line $L$ with negative slope passes through the point $(8,2)$ and positive coordinate axes at points $P$ and $Q$. The absolute minimum value of $OP+OQ$ as $L$ varies, where $O$ is the origin is

Answer 1

Since the slope is negative, let the slope of line be $-m$ where $m$ is positive.

The equation of line is $y-2=-m(x-8)$

$\Rightarrow y+mx=2+8m$

Now the y intercept is $2+8m$ and the x intercept is $8+\frac{2}{m}$

Sum is equal to $10+8m+\frac{2}{m}$

Since, A.M$>$G.M

So, $\Rightarrow \frac{a+b}{2}$$>$$\sqrt{ab}$ for any two positive values of a and b.

Let $a=8m$ and $b=\frac{2}{m}$ then

$\Rightarrow \frac{8m+\frac{2}{m}}{2}>\sqrt{8m\times \frac{2}{m}}$

$\Rightarrow \frac{8m+\frac{2}{m}}{2}>\sqrt{16}$

$\Rightarrow \frac{8m+\frac{2}{m}}{2}>4$

$\Rightarrow 8m+\frac{2}{m}>8$

The minimum value of $8m+\frac{2}{m}$ is $8$. (By AM,GM inequality).

Therefore, the minimum value of $10+8m+\frac{2}{m}$ is $10+8=18$

The minimum value of intercepts is $18$