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Step-by-Step Solution
Step 1: Find the equation of line L1
Line L1 passes through points (1, 2) and (–3, 4). Using the two-point form of a line:
$$
\frac{y - y_1}{x - x_1} \;=\; \frac{y_2 - y_1}{x_2 - x_1}
$$
Substituting $(x_1,\, y_1) = (1,\, 2)$ and $(x_2,\, y_2) = (-3,\, 4)$, we get:
$$
\frac{y - 2}{x - 1} \;=\; \frac{4 - 2}{-3 - 1}
\quad\Longrightarrow\quad
\frac{y - 2}{x - 1} \;=\; \frac{2}{-4} \;=\; -\frac{1}{2}.
$$
Hence,
$$
y - 2 = -\tfrac{1}{2}\,(x - 1).
$$
Simplifying,
$$
2y - 4 = -x + 1
\quad\Longrightarrow\quad
x + 2y = 5.
$$
This is the equation of line L1.
Step 2: Use the fact that (h, k) lies on L1
Since (h, k) lies on L1, it must satisfy the equation $x + 2y = 5$. Therefore:
$$
h + 2k = 5
\quad\quad \text{(Equation 1)}.
$$
Step 3: Find the equation of line L2, perpendicular to L1
The slope of L1 is $-\tfrac{1}{2}$. For L2 to be perpendicular to L1, its slope must be the negative reciprocal of $-\tfrac{1}{2}$, which is 2. Thus, the equation of L2 can be written (in slope–intercept form) as:
$$
y = 2x + c,
$$
or equivalently
$$
2x - y = \lambda.
$$
We determine $\lambda$ by using the point (4, 3), which lies on L2:
$$
2(4) - 3 = \lambda
\quad\Longrightarrow\quad
8 - 3 = 5.
$$
Hence, $\lambda = 5$, giving the equation of L2 as:
$$
2x - y = 5.
$$
Step 4: Use the fact that (h, k) also lies on L2
Since (h, k) is on L2, it must satisfy $2x - y = 5$. Substituting $(x, y) = (h, k)$:
$$
2h - k = 5
\quad\quad \text{(Equation 2)}.
$$
Step 5: Solve for h and k
We have two simultaneous equations:
$$
\text{(1) } h + 2k = 5,
$$
$$
\text{(2) } 2h - k = 5.
$$
From equation (1):
$$
h = 5 - 2k.
$$
Substitute $h$ into equation (2):
$$
2(5 - 2k) - k = 5
\quad\Longrightarrow\quad
10 - 4k - k = 5
\quad\Longrightarrow\quad
10 - 5k = 5
\quad\Longrightarrow\quad
5k = 5
\quad\Longrightarrow\quad
k = 1.
$$
Then $h = 5 - 2(1) = 5 - 2 = 3.$
Step 6: Find the ratio k/h
We have $h = 3$ and $k = 1$, so
$$
\frac{k}{h} = \frac{1}{3}.
$$
Final Answer
$$
\frac{k}{h} = \frac{1}{3}.
$$
Hence, the correct option is $\tfrac{1}{3}$.
