Area of Triangle from Three Intercepts

The equation of the line in intercept form is $$\frac{x}{a}+\frac{y}{b}=1.$$

Since the line passes through $$P(1,2),$$

we have $$\frac1a+\frac2b=1.$$

The area of $$\triangle OAB$$ is $$\Delta=\frac12ab.$$

Using AM-GM,

$$\frac{\frac1a+\frac2b}{2}\ge\sqrt{\frac2{ab}}.$$

Since

$$\frac1a+\frac2b=1,$$

we get

$$\frac12\ge\sqrt{\frac2{ab}}.$$

Therefore,

$$ab\ge8.$$

Hence the minimum area occurs when equality holds in AM-GM, i.e.,

$$\frac1a=\frac2b.$$

Using

$$\frac1a+\frac2b=1,$$

we obtain

$$\frac2a=1\implies a=2.$$

Also,

$$\frac2b=\frac12\implies b=4.$$

Thus the required line is $$\frac{x}{2}+\frac{y}{4}=1,$$ or $$2x+y-4=0.$$

Let

$$M(x,y)$$ be the foot of the perpendicular from the origin to this line.

For the line $$Ax+By+C=0,$$ the foot of the perpendicular from $$(0,0)$$ is

$$\left(\frac{-AC}{A^2+B^2},\frac{-BC}{A^2+B^2}\right).$$

Here

$$A=2,\quad B=1,\quad C=-4.$$

Therefore,

$$M=\left(\frac{-2(-4)}{2^2+1^2},\frac{-1(-4)}{2^2+1^2}\right).$$

$$M=\left(\frac85,\frac45\right).$$

Therefore,Required value=$$5\left(\ \frac{\ 12}{5}\right)$$ = $$12$$