Question

Find the locus of the midpoint of a rod of length 'l' which slides on the lines x + y= 0 and and x - y = 0(each endpoint lies on one of the lines)

• A. The locus of the midpoint is the circle with the equation $x^2 + y^2 = \frac{l^2}{4}$.
• B. The locus of the midpoint is the circle with the equation $x^2 + y^2 = l^2$.
• C. The locus of the midpoint is the line $x+y=0$.
• D. The locus of the midpoint is the line $x-y=0$.

Answer 1

Answer: (A)

The locus of the midpoint is the circle with the equation $x^2 + y^2 = \frac{l^2}{4}$.

Answer 2

Let the two lines be $L_1: x + y = 0$ and $L_2: x - y = 0$. Let the endpoints of the rod of length $l$ be A and B. Let A lie on $L_1$, so its coordinates can be represented as $(a, -a)$. Let B lie on $L_2$, so its coordinates can be represented as $(b, b)$.

The distance between A and B is $l$. Using the distance formula: $AB^2 = (b - a)^2 + (b - (-a))^2 = l^2$ $(b - a)^2 + (b + a)^2 = l^2$ $(b^2 - 2ab + a^2) + (b^2 + 2ab + a^2) = l^2$ $2a^2 + 2b^2 = l^2$ $a^2 + b^2 = \frac{l^2}{2}$

Let the midpoint of the rod be $M(h, k)$. The coordinates of the midpoint are given by: $h = \frac{a + b}{2} \implies 2h = a + b$ $k = \frac{-a + b}{2} \implies 2k = b - a$

We have a system of equations:

1. $a + b = 2h$
2. $b - a = 2k$

Adding (1) and (2): $2b = 2h + 2k \implies b = h + k$ Subtracting (2) from (1): $2a = 2h - 2k \implies a = h - k$

Substitute these expressions for $a$ and $b$ into the equation $a^2 + b^2 = \frac{l^2}{2}$: $(h - k)^2 + (h + k)^2 = \frac{l^2}{2}$ $(h^2 - 2hk + k^2) + (h^2 + 2hk + k^2) = \frac{l^2}{2}$ $2h^2 + 2k^2 = \frac{l^2}{2}$ $h^2 + k^2 = \frac{l^2}{4}$

To find the locus, we replace $(h, k)$ with $(x, y)$: $x^2 + y^2 = \frac{l^2}{4}$

This is the equation of a circle centered at the origin $(0, 0)$ with a radius of $\frac{l}{2}$.