Animated Solution for Mathematics - Straight Lines: If $(1,2),(4,6),(5,7)$ and $(a,b)$ are the vertices of a parallelogram $PQRS$, then

Imagine you are standing on a vast coordinate plane, looking at four points that form a perfect parallelogram. You know three of them: $P(1,2)$, $Q(4,6)$, and $R(5,7)$.

The fourth point, $S(a,b)$, is playing hide-and-seek. While many students immediately reach for the distance formula or slope calculations, there is a much more elegant path.

Geometry is often about finding the right perspective. For a parallelogram, that perspective is the center.

The defining characteristic of a parallelogram is that its diagonals bisect each other. This means if you draw a line from $P$ to $R$ and another from $Q$ to $S$, they will intersect at a single point, $M$, which is the midpoint of both.

This is our golden key. If we find the midpoint of $PR$, we have found the midpoint of $QS$.

Let's calculate the midpoint $M$ of $PR$ using the standard formula:

Substituting our values for $P(1,2)$ and $R(5,7)$, we get:

Simplifying this, we find:

This point $(3,4.5)$ is the heart of our parallelogram.

Now, we turn our attention to the diagonal $QS$. We know $Q$ is at $(4,6)$ and $S$ is at $(a,b)$.

The midpoint of $QS$ must also be $(3,4.5)$. So, we set up the equation:

Now, we simply equate the components. For the $x$-coordinate:

Multiplying by $2$, we get $4+a=6$, which leads us directly to $a=2$.

For the $y$-coordinate:

Multiplying by $2$, we get $6+b=9$, which gives us $b=3$.

Just like that, the mystery of the fourth vertex is solved. The coordinates of $S$ are $(2,3)$.

By using the symmetry of the diagonals, we avoided messy square roots and complex algebra. This is the beauty of coordinate geometry—when you understand the underlying properties, the math flows effortlessly.

Keep this tool in your arsenal; whenever you see a parallelogram, think of the diagonals, and the solution will reveal itself.