Key Concepts and Formulas

• Slope of a Line: Given two points $(x_1, y_1)$ and $(x_2, y_2)$ on a line, the slope $m$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Point-Slope Form: The equation of a line with slope $m$ passing through the point $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$.
• Perpendicular Lines: Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1 m_2 = -1$.

Step-by-Step Solution

Step 1: Find the slope of line $L_1$

We are given that $L_1$ passes through the points $(1, 2)$ and $(-3, 4)$.

• Why this step? We need to determine the slope of $L_1$ to eventually find its equation and the slope of the perpendicular line $L_2$.

Using the slope formula with $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (-3, 4)$: $m_1 = \frac{4 - 2}{-3 - 1} = \frac{2}{-4} = -\frac{1}{2}$

Step 2: Find the equation of line $L_1$

We know the slope of $L_1$ is $-\frac{1}{2}$ and it passes through $(1,2)$.

• Why this step? We need the equation of $L_1$ to use the fact that $(h,k)$ lies on it.

Using the point-slope form: $y - 2 = -\frac{1}{2}(x - 1)$ Multiplying both sides by 2: $2(y - 2) = -(x - 1)$ $2y - 4 = -x + 1$ $x + 2y = 5$

Step 3: Use the fact that $(h, k)$ lies on $L_1$

Since $(h, k)$ lies on $L_1$, it must satisfy the equation of $L_1$.

• Why this step? This will give us our first equation involving $h$ and $k$.

Substituting $x = h$ and $y = k$ into the equation $x + 2y = 5$: $h + 2k = 5 \quad \text{(Equation 1)}$

Step 4: Find the slope of line $L_2$

Line $L_2$ is perpendicular to $L_1$.

• Why this step? We need to find the slope of $L_2$ to eventually find its equation.

Since $L_1 \perp L_2$, $m_1 m_2 = -1$. We know $m_1 = -\frac{1}{2}$. Therefore, $m_2 = \frac{-1}{m_1} = \frac{-1}{-\frac{1}{2}} = 2$.

Step 5: Find the equation of line $L_2$

Line $L_2$ passes through the point $(4, 3)$ and has a slope of $2$.

• Why this step? We need the equation of $L_2$ to use the fact that $(h,k)$ lies on it.

Using the point-slope form: $y - 3 = 2(x - 4)$ $y - 3 = 2x - 8$ $2x - y = 5$

Step 6: Use the fact that $(h, k)$ lies on $L_2$

Since $(h, k)$ lies on $L_2$, it must satisfy the equation of $L_2$.

• Why this step? This will give us our second equation involving $h$ and $k$.

Substituting $x = h$ and $y = k$ into the equation $2x - y = 5$: $2h - k = 5 \quad \text{(Equation 2)}$

Step 7: Solve the system of equations for $h$ and $k$

We have the following system of equations: $h + 2k = 5 \quad \text{(Equation 1)}$ $2h - k = 5 \quad \text{(Equation 2)}$

• Why this step? We need to find the values of $h$ and $k$.

Multiply Equation 2 by 2: $4h - 2k = 10 \quad \text{(Equation 3)}$ Add Equation 1 and Equation 3: $(h + 2k) + (4h - 2k) = 5 + 10$ $5h = 15$ $h = 3$ Substitute $h = 3$ into Equation 1: $3 + 2k = 5$ $2k = 2$ $k = 1$ Therefore, $(h, k) = (3, 1)$.

Step 8: Calculate $\frac{k}{h}$

We need to find the value of $\frac{k}{h}$.

• Why this step? This is the final calculation to answer the question.

$\frac{k}{h} = \frac{1}{3}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating slopes and manipulating equations.
• Perpendicular Slopes: Ensure you take the negative reciprocal of the slope, not just the negative.
• System of Equations: Double-check your work when solving the system of equations to avoid arithmetic errors.

Summary

We found the equation of line $L_1$ using the two-point formula, then used the fact that $(h,k)$ lies on $L_1$ to get our first equation in $h$ and $k$. We then found the equation of the perpendicular line $L_2$ using the given point $(4,3)$ and the perpendicularity condition. Substituting $(h,k)$ into the equation of $L_2$ gave us our second equation in $h$ and $k$. We solved the system of equations to find $h=3$ and $k=1$, and finally calculated $\frac{k}{h} = \frac{1}{3}$.

The final answer is \boxed{1/3}, which corresponds to option (A).