Key Concepts and Formulas

• Equation of a line: The equation of a line can be represented in various forms, including slope-intercept form ($y = mx + c$) and point-slope form ($y - y_1 = m(x - x_1)$).
• Perpendicular lines: If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1 \cdot m_2 = -1$.
• Midpoint formula: The midpoint of a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

Step-by-Step Solution

Step 1: Define a general point on the line $x = 2y$. Let a point on the line $x = 2y$ be $A(2t, t)$, where $t$ is a parameter. This represents any arbitrary point on the line.

Step 2: Define the midpoint of the perpendicular and the point on the line $x = y$. Let the midpoint of the perpendicular from $A(2t, t)$ to the line $x = y$ be $M(h, k)$. Let the foot of the perpendicular from $A$ to the line $x = y$ be $B(x_1, y_1)$. Since B lies on $x=y$, we have $x_1=y_1$.

Step 3: Relate the coordinates of the midpoint to the coordinates of the endpoints. Since $M(h, k)$ is the midpoint of $AB$, we have: $h = \frac{2t + x_1}{2} \quad \text{and} \quad k = \frac{t + y_1}{2}$ Thus, $2h = 2t + x_1$ and $2k = t + y_1$. Since $x_1 = y_1$, we have $x_1 = 2h - 2t \quad \text{and} \quad x_1 = 2k - t$

Step 4: Find the slope of the line $AB$ and use the perpendicularity condition. The slope of the line $AB$ is given by: $m_{AB} = \frac{y_1 - t}{x_1 - 2t} = \frac{x_1 - t}{x_1 - 2t}$ The slope of the line $x = y$ is $1$. Since $AB$ is perpendicular to $x = y$, we have $m_{AB} \cdot 1 = -1$. Therefore, $\frac{x_1 - t}{x_1 - 2t} = -1$ $x_1 - t = -x_1 + 2t$ $2x_1 = 3t$

Step 5: Eliminate the parameter $t$ and $x_1$. We have the following equations: $x_1 = 2h - 2t \quad (1)$ $x_1 = 2k - t \quad (2)$ $2x_1 = 3t \quad (3)$ From (3), we have $x_1 = \frac{3}{2}t$. Substituting this into (1) and (2), we get: $\frac{3}{2}t = 2h - 2t \quad \text{and} \quad \frac{3}{2}t = 2k - t$ From the first equation, we have $\frac{3}{2}t + 2t = 2h$, which gives $\frac{7}{2}t = 2h$, so $t = \frac{4h}{7}$. From the second equation, we have $\frac{3}{2}t + t = 2k$, which gives $\frac{5}{2}t = 2k$, so $t = \frac{4k}{5}$. Equating the two expressions for $t$, we have: $\frac{4h}{7} = \frac{4k}{5}$ $\frac{h}{7} = \frac{k}{5}$ $5h = 7k$ $5h - 7k = 0$

Step 6: Replace $(h, k)$ with $(x, y)$ to obtain the locus. Replacing $h$ with $x$ and $k$ with $y$, we get the equation of the locus: $5x - 7y = 0$

Step 7: Check for Errors

Oops! There's a mistake. Let's rework from Step 5.

From $x_1 = 2h - 2t$ and $x_1 = 2k - t$, we can write $2k - t = 2h - 2t$, so $t = 2h - 2k$. Substituting $t = 2h - 2k$ into $2x_1 = 3t$, we have $x_1 = \frac{3}{2}t = \frac{3}{2}(2h - 2k) = 3h - 3k$.

Now, substitute $x_1 = 3h - 3k$ into $x_1 = 2h - 2t$. Then $3h - 3k = 2h - 2t$, so $h - 3k = -2t$, thus $t = \frac{3k-h}{2}$. Substituting $t = \frac{3k-h}{2}$ into $x_1 = 2k - t$, we have $x_1 = 2k - \frac{3k-h}{2} = \frac{4k - 3k + h}{2} = \frac{h+k}{2}$. Then $3h-3k = \frac{h+k}{2}$, so $6h - 6k = h + k$, which means $5h = 7k$, or $5x = 7y$. This is still wrong.

Going back to the start and trying a different approach: From equations (1) and (2): $2h-2t = 2k-t \implies t = 2h-2k$. The slope of the line joining $(2t, t)$ and $(x_1, y_1)$ is $\frac{t-y_1}{2t-x_1}$. Since $x_1 = y_1$ and the line $x=y$ has slope 1, we must have $\frac{t-x_1}{2t-x_1} = -1$. So $t-x_1 = -2t + x_1 \implies 3t = 2x_1$, thus $x_1 = \frac{3}{2}t$. Since $x_1 = \frac{3}{2}t$, we have $h = \frac{2t+x_1}{2} = \frac{2t+\frac{3}{2}t}{2} = \frac{7}{4}t$ and $k = \frac{t+x_1}{2} = \frac{t+\frac{3}{2}t}{2} = \frac{5}{4}t$. So $t = \frac{4}{7}h$ and $t = \frac{4}{5}k$. Then $\frac{4}{7}h = \frac{4}{5}k$, so $5h = 7k$, or $5x-7y = 0$. Still wrong.

Let's reconsider the perpendicular slope. The slope of the line $x=y$ is 1. Thus the slope of the perpendicular is $-1$. So $\frac{k-t}{h-2t} = -1$, so $k-t = -h+2t \implies 3t = h+k$, so $t = \frac{h+k}{3}$. Also, $h = \frac{2t+x_1}{2}$ and $k = \frac{t+x_1}{2}$, so $2h = 2t+x_1$ and $2k = t+x_1$. Subtracting, we have $2h-2k = t$, so $t = 2h-2k$. Then $\frac{h+k}{3} = 2h-2k$, so $h+k = 6h-6k \implies 5h = 7k$, so $5x-7y=0$. Still getting the wrong answer.

Let A be $(2t, t)$. The line $x=y$ has slope 1. A line perpendicular to $x=y$ has slope $-1$. The equation of the perpendicular line through $(2t, t)$ is $y-t = -1(x-2t)$, so $y = -x + 3t$. The intersection of $x=y$ and $y = -x + 3t$ is $x = -x+3t$, so $2x = 3t$, thus $x = \frac{3t}{2}$. So the point of intersection is $(\frac{3t}{2}, \frac{3t}{2})$. The midpoint of $(2t, t)$ and $(\frac{3t}{2}, \frac{3t}{2})$ is $(\frac{2t+\frac{3t}{2}}{2}, \frac{t+\frac{3t}{2}}{2}) = (\frac{7t}{4}, \frac{5t}{4})$. So $h = \frac{7t}{4}$ and $k = \frac{5t}{4}$. Then $t = \frac{4h}{7}$ and $t = \frac{4k}{5}$. So $\frac{4h}{7} = \frac{4k}{5}$, which means $5h = 7k$, so $5x = 7y$, or $5x-7y = 0$. Still wrong.

Correct Approach: Let the line $x = 2y$ be $L_1$ and the line $x = y$ be $L_2$. Let a point on $L_1$ be $(2t, t)$. The equation of the line perpendicular to $x = y$ and passing through $(2t, t)$ is $y - t = -1(x - 2t)$, i.e., $y = -x + 3t$. The intersection point of $x = y$ and $y = -x + 3t$ is obtained by substituting $x$ for $y$ in the second equation: $x = -x + 3t$, so $2x = 3t$, and $x = \frac{3t}{2}$. Therefore, the point is $(\frac{3t}{2}, \frac{3t}{2})$. Let the midpoint of the perpendicular be $(h, k)$. Then $h = \frac{2t + \frac{3t}{2}}{2} = \frac{7t}{4}$ and $k = \frac{t + \frac{3t}{2}}{2} = \frac{5t}{4}$. Eliminating $t$, we have $t = \frac{4h}{7}$ and $t = \frac{4k}{5}$. Thus, $\frac{4h}{7} = \frac{4k}{5}$, which gives $5h = 7k$. Hence, the locus is $5x - 7y = 0$. Still incorrect.

Another attempt: A point on $x = 2y$ is $(2t, t)$. Let the foot of the perpendicular on $x = y$ be $(a, a)$. The midpoint is $(h, k) = (\frac{2t+a}{2}, \frac{t+a}{2})$. The slope of the line joining $(2t, t)$ and $(a, a)$ is $\frac{a-t}{a-2t}$. Since the line is perpendicular to $x=y$, we have $\frac{a-t}{a-2t} = -1$. So $a-t = -a + 2t$, thus $2a = 3t$, $a = \frac{3t}{2}$. $h = \frac{2t+\frac{3t}{2}}{2} = \frac{7t}{4}$ and $k = \frac{t+\frac{3t}{2}}{2} = \frac{5t}{4}$. Then $t = \frac{4h}{7}$ and $t = \frac{4k}{5}$. So $\frac{4h}{7} = \frac{4k}{5}$, which gives $5h=7k$. So $5x-7y=0$.

Let's try to find an error in the question or answer. Let's try the point (0,0) on the line $x=2y$. The perpendicular to $x=y$ from (0,0) is $y=-x$. Intersection is (0,0). The midpoint is (0,0). Substituting into $5x-7y=0$ works. Substituting into $3x-2y=0$ also works.

Let's try the point (2,1) on the line $x=2y$. The perpendicular to $x=y$ is $y-1 = -1(x-2)$, so $y = -x+3$. The intersection of $x=y$ and $y=-x+3$ is $x=-x+3$, so $2x=3$, $x=3/2$. So the point is $(3/2, 3/2)$. The midpoint is $(\frac{2+3/2}{2}, \frac{1+3/2}{2}) = (\frac{7/2}{2}, \frac{5/2}{2}) = (\frac{7}{4}, \frac{5}{4})$. Plugging this into $3x-2y=0$, we get $3(\frac{7}{4}) - 2(\frac{5}{4}) = \frac{21-10}{4} = \frac{11}{4} \neq 0$. So $3x-2y=0$ is NOT the answer. Plugging this into $5x-7y=0$, we get $5(\frac{7}{4}) - 7(\frac{5}{4}) = \frac{35-35}{4} = 0$.

Final Correct Approach Let a point on $x = 2y$ be $(2t, t)$. Let the midpoint of the perpendicular from $(2t, t)$ to $x = y$ be $(h, k)$. Let the foot of the perpendicular on $x=y$ be $(a, a)$. Then $(h, k) = \left( \frac{2t+a}{2}, \frac{t+a}{2} \right)$. The slope of the perpendicular is $\frac{a-t}{a-2t} = -1$. Thus $a-t = -a+2t$, so $2a = 3t$, and $a = \frac{3}{2}t$. Substituting into the midpoint equations, we have $h = \frac{2t+\frac{3}{2}t}{2} = \frac{\frac{7}{2}t}{2} = \frac{7}{4}t$, and $k = \frac{t+\frac{3}{2}t}{2} = \frac{\frac{5}{2}t}{2} = \frac{5}{4}t$. Thus, $t = \frac{4h}{7}$ and $t = \frac{4k}{5}$. Equating the two expressions for $t$, we have $\frac{4h}{7} = \frac{4k}{5}$, so $5h = 7k$, which implies $5x = 7y$. Therefore, the locus is $5x - 7y = 0$.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when calculating slopes and applying the perpendicularity condition.
• Parameter Elimination: Ensure you eliminate the parameter correctly to obtain the locus equation.
• Midpoint Formula: Double-check the application of the midpoint formula.

Summary

We found a general point on the line $x = 2y$ and then found the foot of the perpendicular to the line $x = y$. We then used the midpoint formula to relate the coordinates of the midpoint to the coordinates of the points on the lines. Finally, we eliminated the parameter to find the locus of the midpoint. After several attempts, the derivation consistently yields the result $5x - 7y = 0$. The given correct answer is incorrect.

Final Answer The final answer is \boxed{5x-7y=0}. However, the correct answer is NOT among the options (A), (B), (C), or (D). There seems to be an error in the options provided.