Let $$L$$ be the line passing through the point $$P(1, 2)$$ such that its intercepted segment between the co-ordinate axes is bisected at $$P$$. If $$L_1$$ is the line perpendicular to $$L$$ and passing through the point $$(-2, 1)$$, then the point of intersection of $$L$$ and $$L_1$$ is

We are told that the required line $$L$$ passes through the fixed point $$P(1,2)$$ and that the segment cut by this line on the co-ordinate axes is bisected at $$P$$.

To express an ordinary straight line in terms of its intercepts, we use the axis-intercept form (state the formula first):

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

where $$a$$ is the $$x$$-intercept (the point $$(a,0)$$ on the $$x$$-axis) and $$b$$ is the $$y$$-intercept (the point $$(0,b)$$ on the $$y$$-axis).

The midpoint of this intercepted segment with endpoints $$(a,0)$$ and $$(0,b)$$ is obtained from the midpoint formula $$\bigl(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\bigr).$$

So the midpoint is

$$\left(\frac{a+0}{2},\frac{0+b}{2}\right)=\left(\frac{a}{2},\frac{b}{2}\right).$$

But we are told that this midpoint is exactly the point $$P(1,2).$$ Hence we can write two separate equalities, equating corresponding co-ordinates:

$$\frac{a}{2}=1 \quad\text{and}\quad \frac{b}{2}=2.$$

From the first equality we immediately get

$$a = 2,$$

and from the second equality we get

$$b = 4.$$

Now substitute $$a=2$$ and $$b=4$$ back into the intercept form of the line:

$$\frac{x}{2} + \frac{y}{4} = 1.$$

To put this in a simpler linear form, multiply every term by $$4$$:

$$4\left(\frac{x}{2}\right) + 4\left(\frac{y}{4}\right) = 4(1).$$

This simplifies step by step to

$$2x + y = 4.$$

So the explicit equation of line $$L$$ is

$$2x + y = 4,$$

which can also be rearranged as

$$y = 4 - 2x.$$

From $$y = 4 - 2x$$ we identify the slope of $$L$$. Writing the general slope-intercept form $$y = mx + c,$$ we compare and see that

$$m_1 = -2.$$

Next, we are asked for a line $$L_1$$ that is perpendicular to $$L$$ and that also passes through the point $$(-2,1).$$

If two lines are perpendicular, the product of their slopes equals $$-1$$. Hence, denoting by $$m_2$$ the slope of $$L_1$$, we use the perpendicular-slopes condition:

$$m_1 \, m_2 = -1.$$

Substituting $$m_1 = -2,$$ we obtain

$$(-2)\, m_2 = -1.$$

Dividing both sides by $$-2$$, we get

$$m_2 = \frac{1}{2}.$$

Thus the slope of $$L_1$$ is $$\frac12$$, and we know that the line must pass through the given point $$(-2,1).$$ We now write the equation of $$L_1$$ in point-slope form (state the formula):

$$y - y_0 = m(x - x_0),$$

where $$(x_0,y_0)=(-2,1)$$. Substituting these values and $$m = \tfrac12$$, we get

$$y - 1 = \frac12 \bigl(x - (-2)\bigr).$$

Since $$x - (-2) = x + 2,$$ the equation becomes

$$y - 1 = \frac12 (x + 2).$$

To make later substitution easy, expand the right-hand side:

$$y - 1 = \frac{x}{2} + 1.$$

Now add $$1$$ to both sides:

$$y = \frac{x}{2} + 2.$$

So the explicit equation of $$L_1$$ is

$$y = \frac{x}{2} + 2.$$

We must now find the point of intersection of the two lines. This point must satisfy both

$$y = 4 - 2x \quad\text{and}\quad y = \frac{x}{2} + 2.$$

We therefore set the two right-hand sides equal:

$$4 - 2x = \frac{x}{2} + 2.$$

To clear the fraction and make the algebra straightforward, multiply every term by $$2$$:

$$2(4) - 2(2x) = 2\left(\frac{x}{2}\right) + 2(2).$$

This simplifies to

$$8 - 4x = x + 4.$$

Now gather all the $$x$$-terms on one side and the constants on the other side. First add $$4x$$ to both sides:

$$8 = x + 4 + 4x.$$

That gives

$$8 = 5x + 4.$$

Next, subtract $$4$$ from both sides to isolate the term with $$x$$:

$$8 - 4 = 5x.$$

Hence

$$4 = 5x.$$

Divide by $$5$$ to obtain $$x$$:

$$x = \frac{4}{5}.$$

With this value of $$x$$, we substitute back into either of the two line equations to get $$y$$. Choosing the simpler $$y = 4 - 2x$$, we write

$$y = 4 - 2\left(\frac{4}{5}\right).$$

Compute the product inside the parentheses first:

$$2\left(\frac{4}{5}\right) = \frac{8}{5}.$$

Thus

$$y = 4 - \frac{8}{5}.$$

Rewrite $$4$$ as $$\frac{20}{5}$$ so we share a common denominator, and subtract:

$$y = \frac{20}{5} - \frac{8}{5} = \frac{12}{5}.$$

We have therefore obtained the intersection point

$$\left(\frac{4}{5},\frac{12}{5}\right).$$

This matches option B in the list provided. Hence, the correct answer is Option B.