Question

Let the line $\frac{x}{4} + \frac{y}{2} = 1$ meet the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line $x + y = 1$. Let the point P lie on the line $x + y = 1$ such that $\Delta ABP$ is an isosceles triangle with $AP = BP$. Then the distance between M' and P is:
 

• A. $\frac{\sqrt{7}}{3}$
• B. $\frac{2\sqrt{5}}{3}$
• C. $\frac{2\sqrt{7}}{3}$
• D. $\frac{\sqrt{5}}{3}$

Hint:

To find distances between points or reflections, use the midpoint formula and reflection formulas for accurate calculations.

Answer 1

The Correct Option is B

The line $\frac{x}{4} + \frac{y}{2} = 1$ meets the x-axis at point A (where $y = 0$) and the y-axis at point B (where $x = 0$). Solving for these points:
For point A: Substituting $y = 0$ in the equation of the line: \[ \frac{x}{4} = 1 \implies x = 4 \] Thus, A = (4, 0). For point B: Substituting $x = 0$ in the equation of the line: \[ \frac{y}{2} = 1 \implies y = 2 \] Thus, B = (0, 2). The midpoint M of AB is the average of the coordinates of A and B: \[ M = \left( \frac{4+0}{2}, \frac{0+2}{2} \right) = (2, 1) \] The point M' is the reflection of M across the line $x + y = 1$. Using the formula for the reflection of a point across a line, we find that the coordinates of M' are $(2, 3)$. The line $x + y = 1$ intersects the line $x + y = 3$ (the line containing P). Since $\Delta ABP$ is isosceles, we calculate the distance between M' and P, which is $\frac{2\sqrt{5}}{3}$.