Animated Solution for Mathematics - Straight Lines: The locus of mid points 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. You see two lines stretching out before you: $L_1$, defined by the equation $x=2y$, and $L_2$, the classic $x=y$. These lines intersect at the origin, creating a beautiful, sharp angle.

Now, imagine a point $P$ sliding along $L_1$. As it moves, you drop a perpendicular line from $P$ down to $L_2$, hitting it at a point $Q$. Your task is to track the midpoint $R$ of this segment $PQ$.

This is the essence of a locus problem—finding the hidden path traced by a point under a geometric constraint.

To solve this, we must first understand the geometry of our perpendicular. The line $L_2$ is $y=x$, which has a slope of $1$.

Any line perpendicular to this must have a slope $m$ such that the product of the slopes is $-1$. Thus, $m\cdot 1=-1$, giving us $m=-1$.

This is our key! Any line perpendicular to $L_2$ must take the form $y=-x+c$, or more elegantly, $x+y=c$. Here, $c$ is our 'parameter'—a variable that changes as $P$ moves, effectively labeling every possible perpendicular line we could draw.

Now, let us find the coordinates of $P$ and $Q$. Since $Q$ is the intersection of our perpendicular line $x+y=c$ and the line $L_2$ ($x=y$), we substitute $y=x$ into the perpendicular equation: $x+x=c$, which gives $x=\frac{c}{2}$.

Therefore, the coordinates of $Q$ are:

Next, we find $P$. $P$ is the intersection of the same perpendicular line $x+y=c$ and the line $L_1$ ($x=2y$). Substituting $x=2y$ into $x+y=c$, we get $2y+y=c$, so $3y=c$, which means $y=\frac{c}{3}$.

Since $x=2y$, we find $x=\frac{2c}{3}$. Thus, the coordinates of $P$ are:

We define the midpoint $R$ as $(h,k)$. Using the midpoint formula, we know that $h=\frac{x_P+x_Q}{2}$ and $k=\frac{y_P+y_Q}{2}$.

Substituting our expressions for $P$ and $Q$ into these formulas, we get:

Look at the elegance of these results! We have expressed the coordinates of our moving midpoint $R$ entirely in terms of the parameter $c$.

To find the locus, we must eliminate $c$. From our equations, we see that $c=\frac{12h}{7}$ and $c=\frac{12k}{5}$.

Since both expressions equal $c$, they must be equal to each other:

Canceling the $12$ from both sides and cross-multiplying, we arrive at $5h=7k$. Replacing $(h,k)$ with the general coordinates $(x,y)$, we find the equation of the locus:

This is the path of the midpoint! It is a straight line passing through the origin, perfectly bisecting the space between the original lines. You have successfully navigated the geometry, mastered the parameter, and uncovered the hidden path.