Without using distance formula, show that points $(-2, -1)$, $(4, 0)$, $(3, 3)$ and $(-3, 2)$ are the vertices of a parallelogram.

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> [!TLDR]
> We can prove these four points form a parallelogram by showing that the midpoints of the two diagonals coincide — this is the defining property of a parallelogram, and it avoids the distance formula entirely. The midpoint of diagonal $AC$ is $(0.5, 1)$ and the midpoint of diagonal $BD$ is also $(0.5, 1)$, so the points are indeed vertices of a parallelogram.

The problem asks us to show that four given points are the vertices of a parallelogram — but without using the distance formula. That’s a deliberate constraint. The distance formula is the usual way to check that opposite sides are equal, but there’s a much cleaner approach using midpoints.

**The key idea:** In any parallelogram, the diagonals bisect each other. That means the midpoint of one diagonal is exactly the same point as the midpoint of the other diagonal. So if we label the four points as $A$, $B$, $C$, and $D$ in order, we just need to check whether the midpoint of $AC$ equals the midpoint of $BD$.

This is faster, simpler, and uses only the midpoint formula — which is just averaging coordinates. No square roots, no distances.

Let’s label the points:
- $A = (-2, -1)$
- $B = (4, 0)$
- $C = (3, 3)$
- $D = (-3, 2)$

We’ll assume the vertices are taken in order $A 	o B 	o C 	o D$. That means the diagonals are $AC$ and $BD$.

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1. **Find the midpoint of diagonal $AC$**  
   The midpoint formula: for two points $(x_1, y_1)$ and $(x_2, y_2)$, the midpoint is  
   $$ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} ight) $$  
   For $A(-2, -1)$ and $C(3, 3)$:  
   $$ 	ext{Midpoint of } AC = \left( \frac{-2 + 3}{2}, \frac{-1 + 3}{2} ight) = \left( \frac{1}{2}, \frac{2}{2} ight) = (0.5, 1) $$

2. **Find the midpoint of diagonal $BD$**  
   For $B(4, 0)$ and $D(-3, 2)$:  
   $$ 	ext{Midpoint of } BD = \left( \frac{4 + (-3)}{2}, \frac{0 + 2}{2} ight) = \left( \frac{1}{2}, \frac{2}{2} ight) = (0.5, 1) $$

3. **Compare the two midpoints**  
   Both are $(0.5, 1)$. They are identical.

Quadrant	x-sign	y-sign	Example
I	$+$	$+$	$(2, 3)$
II	$-$	$+$	$(-1, 4)$
III	$-$	$-$	$(-3, -2)$
IV	$+$	$-$	$(5, -1)$

Q.Without using distance formula, show that points $(-2, -1)$ , $(4, 0)$ , $(3, 3)$ and $(-3, 2)$ are the vertices of a parallelogram.

Concept understanding — Coordinate Geometry

Coordinate Geometry: Where Algebra Meets Geometry

The Big Idea

The Coordinate System

A Concrete Example

The Four Quadrants

Why This Matters

The Precise Statement

Q.Without using distance formula, show that points (−2,−1)(-2, -1)(−2,−1), (4,0)(4, 0)(4,0), (3,3)(3, 3)(3,3) and (−3,2)(-3, 2)(−3,2) are the vertices of a parallelogram.

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Q.Without using distance formula, show that points $(-2, -1)$ , $(4, 0)$ , $(3, 3)$ and $(-3, 2)$ are the vertices of a parallelogram.

Unlock everything free for 14 days