Animated Solution for Mathematics - Straight Lines: The locus of the mid-point of the perpendiculars drawn from points on the line, $x=2y$ to the line $x=y$ is:

Imagine you are standing on a coordinate plane with two lines, $L_1:x=2y$ and $L_2:x=y$. A point $P$ slides along $L_1$, and from this point, you drop a perpendicular line segment to $L_2$.

As $P$ moves, this segment moves, and its midpoint traces a path. This path is the geometric signature of the motion, known as the locus.

Parametric Elegance

To capture the motion, we use parameters. Since $P$ lies on $x=2y$, we define its coordinates as $P(2\alpha ,\alpha )$.

Now, consider the foot of the perpendicular, $Q$, which lies on $x=y$. Since any point on this line has equal coordinates, we define $Q(\beta ,\beta )$.

The Geometric Soul

The condition that $PQ$ is perpendicular to $L_2$ is the heartbeat of this problem. The slope of $L_2$ is $1$.

Therefore, the slope of the segment $PQ$ must be $-1$, because the product of perpendicular slopes is $-1$. We write this as:

Solving this, we find $\beta -\alpha =-\beta +2\alpha$, which simplifies to $2\beta =3\alpha$, or:

We have successfully linked the two parameters.

The Midpoint Journey

Now, we define the midpoint $M(h,k)$. Using the midpoint formula, $h=\frac{x_P+x_Q}{2}$ and $k=\frac{y_P+y_Q}{2}$.

Substituting our values, we get:

The Final Elimination

To find the locus, we must eliminate $\alpha$. By dividing $h$ by $k$, we get:

This leads us to $5h=7k$. Replacing $(h,k)$ with $(x,y)$, we arrive at the final equation:

This is the path of the midpoint. It is a straight line, elegant and precise, just like the logic that brought us here.