Question

A variable drawn through a fixed point cuts axes at $A$ and $B$. The locus of the midpoint of $AB$ is

Answer 1

y_0 x + x_0 y = 2xy

Answer 2

Let the fixed point be $(x_0, y_0)$.
Let the variable line passing through $(x_0, y_0)$ cut the x-axis at $A(a, 0)$ and the y-axis at $B(0, b)$.
The equation of the line passing through $A(a, 0)$ and $B(0, b)$ can be written in the intercept form as $\frac{x}{a} + \frac{y}{b} = 1$.
Since the line passes through the fixed point $(x_0, y_0)$, these coordinates must satisfy the equation of the line:
$\frac{x_0}{a} + \frac{y_0}{b} = 1$.

Let $M(h, k)$ be the midpoint of the line segment AB.
The coordinates of the midpoint are given by the average of the coordinates of the endpoints:
$h = \frac{a + 0}{2} = \frac{a}{2} \implies a = 2h$.
$k = \frac{0 + b}{2} = \frac{b}{2} \implies b = 2k$.

Substitute the expressions for $a$ and $b$ in terms of $h$ and $k$ into the equation involving the fixed point:
$\frac{x_0}{2h} + \frac{y_0}{2k} = 1$.

To find the locus of the midpoint $M(h, k)$, we replace $h$ with $x$ and $k$ with $y$:
$\frac{x_0}{2x} + \frac{y_0}{2y} = 1$.

This is the equation of the locus. We can simplify it by multiplying the entire equation by $2xy$:
$x_0 y + y_0 x = 2xy$.

Rearranging the terms, we get: $y_0 x + x_0 y - 2xy = 0$.

This equation represents the locus of the midpoint of AB for a line passing through the fixed point $(x_0, y_0)$. If the fixed point is specified, say $(1, 1)$ as in the similar question, the locus becomes $1 \cdot x + 1 \cdot y - 2xy = 0$, or $x + y - 2xy = 0$.