Equation of Line: Intercept Form

Let the equation of the line be

$$\frac{x}{a}+\frac{y}{b}=1$$

where

$$a=OA$$

and

$$b=OB$$

Since the line passes through

$$(4,9)$$

we have

$$\frac{4}{a}+\frac{9}{b}=1$$

We need to minimize

$$a+b$$

Now,

$$a+b=(a+b)\left(\frac{4}{a}+\frac{9}{b}\right)$$

Expanding,

$$a+b=\frac{4a}{a}+\frac{9a}{b}+\frac{4b}{a}+\frac{9b}{b}$$

$$=13+\frac{9a}{b}+\frac{4b}{a}$$

Applying AM-GM,

$$\frac{\frac{9a}{b}+\frac{4b}{a}}{2}\ge\sqrt{\frac{9a}{b}\cdot\frac{4b}{a}}$$

$$=\sqrt{36}$$

$$=6$$

Therefore,

$$\frac{9a}{b}+\frac{4b}{a}\ge12$$

Hence,

$$a+b\ge13+12$$

$$=25$$

Equality holds when

$$\frac{9a}{b}=\frac{4b}{a}$$

$$9a^2=4b^2$$

$$3a=2b$$

Thus the minimum value of

$$a+b$$

is

$$25$$

1. Determine the vertices and area of the initial triangle:

The boundary lines are given as $$2x + y = 4$$, the positive x-axis ($$y = 0$$), and the positive y-axis ($$x = 0$$).

- To find the x-intercept, we set $$y = 0 \implies 2x = 4 \implies x = 2$$. This gives vertex $$(2, 0)$$.

- To find the y-intercept, we set $$x = 0 \implies y = 4$$. This gives vertex $$(0, 4)$$.

The third vertex sits at the origin $$(0, 0)$$.

The total area of this right-angled triangle is:

$$\text{Total Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4$$

2. Set up the intercepts of the bisecting line:

The line cuts the area exactly in half, meaning the smaller triangle it forms with the origin must have an area of:

$$\text{New Area} = \frac{4}{2} = 2$$

Let the area-bisecting line make an x-intercept at $$x_0$$ and a y-intercept at $$y_0$$. The area equation gives:

$$\frac{1}{2} \cdot x_0 \cdot y_0 = 2 \implies x_0 y_0 = 4$$

3. Use the point constraint to find the intercepts:

The equation of the line written in its standard intercept form is:

$$\frac{x}{x_0} + \frac{y}{y_0} = 1$$

Since the line is explicitly stated to pass through the point $$(1, 1)$$, we substitute these coordinates into the equation:

$$\frac{1}{x_0} + \frac{1}{y_0} = 1 \implies \frac{x_0 + y_0}{x_0 y_0} = 1$$

Substitute our known area product value $$x_0 y_0 = 4$$ into this relation:

$$\frac{x_0 + y_0}{4} = 1 \implies x_0 + y_0 = 4$$

4. Calculate the required absolute sum:

The problem asks for the absolute value of the sum of its intercepts on the coordinate axes, which is:

$$|x_0 + y_0| = |4| = 4$$

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=$$5\left(\ \frac{\ 12}{5}\right)$$ = $$12$$