Three consecutive vertices of a parallelogram ABCD are $A(6,-2,4)$, $B(2,4,-8)$, $C(-2,2,4)$. Find the coordinates of the fourth vertex. [Hint: Diagonals of a parallelogram have the same mid-point.]

Three consecutive vertices of a parallelogram ABCD are A(6,-2,4), B(2,4,-8), C(-2,2,4). Find the coordinates o

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A(6,-2,4)

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Concept understanding — 3D Coordinate Geometry

3D Coordinate Geometry

You already know 2D coordinate geometry — the $xy$ -plane where every point is described by two numbers $(x, y)$ . Now imagine lifting that plane into the air. That is three-dimensional geometry.

The Intuition: Three Numbers, One Point

In the real world you rarely locate something with just two numbers. To describe where a book sits on a shelf you might say: "third shelf up, fourth book from the left, and it is the one nearest the wall." That is three pieces of information — height, sideways position, and depth.

In 3D coordinate geometry we do exactly this. We keep the familiar $x$ and $y$ axes (which define a flat floor) and add a third axis — the $z$ -axis — pointing straight up. Every point in space now needs three numbers: $(x, y, z)$ .

Note

The three axes are mutually perpendicular. Picture the corner of a room: two floor edges give the $x$ - and $y$ -axes, and the vertical edge where the walls meet gives the $z$ -axis.

The Precise Statement

Definition: A rectangular 3D coordinate system consists of three mutually perpendicular number lines — the $x$ -axis, $y$ -axis and $z$ -axis — meeting at a common point, the origin $O(0, 0, 0)$ . Any point $P$ in space is uniquely represented by an ordered triple $(x, y, z)$ , where:

$x$ = signed distance from the $yz$ -plane,
$y$ = signed distance from the $zx$ -plane,
$z$ = signed distance from the $xy$ -plane.

P = (x,\, y,\, z)

How to Read a 3D Point

Take the point $A(2, -3, 4)$ . Start at the origin. Move 2 units along the $x$ -axis. From there move $-3$ units parallel to the $y$ -axis (backward, because it is negative). From that spot move 4 units parallel to the $z$ -axis (upward). You have reached $A$ .

Watch out

The order matters absolutely. $(2, -3, 4)$ is not the same point as $(2, 4, -3)$ . Always follow the sequence: $x$ first, then $y$ , then $z$ .

The Three Coordinate Planes

Each pair of axes determines a plane:

Plane	Equation	Description
$xy$ -plane	$z = 0$	the floor — all points with zero height
$yz$ -plane	$x = 0$	one wall — all points with zero $x$
$zx$ -plane	$y = 0$	the other wall — all points with zero $y$

These three planes cut space into 8 octants (the 3D analogue of the four quadrants of the plane). The first octant is where $x > 0$ , $y > 0$ and $z > 0$ .

Distance Between Two Points

This is the natural extension of the 2D distance formula. For $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ :

PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

It is just the diagonal of a rectangular box whose edges are the differences in $x$ , $y$ and $z$ . The distance of $P$ from the origin is the special case $OP = \sqrt{x_1^2 + y_1^2 + z_1^2}$ .

Section Formula (Internal Division)

If $R$ divides the segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ internally in the ratio $m : n$ , then: …

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The diagonals of a parallelogram bisect each other, so their midpoints coincide. Using this property with vertices $A$ , $B$ , $C$ , we find the fourth vertex $D$ at $(2, -4, 16)$ .

The key insight here is that a parallelogram's diagonals always bisect each other. This means the midpoint of diagonal $AC$ must equal the midpoint of diagonal $BD$ . Once we know three consecutive vertices, this single condition determines the fourth vertex uniquely.

Why the midpoint condition works

In any parallelogram, opposite sides are parallel and equal. If we label the vertices in order as $A$ , $B$ , $C$ , $D$ , then the diagonals are $AC$ and $BD$ . The defining property of a parallelogram guarantees that these diagonals meet at their common midpoint. So if we know $A$ , $B$ , and $C$ , we can find $D$ by enforcing:

\text{Midpoint of } AC = \text{Midpoint of } BD

Let's work through this step by step.

Solution

1. Identify the diagonal containing the known vertices

We have three consecutive vertices: $A(6, -2, 4)$ , $B(2, 4, -8)$ , and $C(-2, 2, 4)$ . Since they are consecutive, the vertices are ordered around the parallelogram. The diagonals are $AC$ and $BD$ , where $D$ is the unknown fourth vertex.

2. Calculate the midpoint of diagonal $AC$

The midpoint $M$ of a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)

For diagonal $AC$ :

M_{AC} = \left( \frac{6 + (-2)}{2}, \frac{-2 + 2}{2}, \frac{4 + 4}{2} \right) = \left( \frac{4}{2}, \frac{0}{2}, \frac{8}{2} \right) = (2, 0, 4)

3. Set up the midpoint equation for diagonal $BD$

Let the unknown vertex $D$ have coordinates $(x, y, z)$ . The midpoint of diagonal $BD$ is:

$M_{BD} = \left( \frac{2 + x}{2}, \frac{4 + y}{2}, \frac{-8 + z}{2} \right)$ …

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Q.Three consecutive vertices of a parallelogram ABCD are A(6,−2,4)A(6,-2,4)A(6,−2,4), B(2,4,−8)B(2,4,-8)B(2,4,−8), C(−2,2,4)C(-2,2,4)C(−2,2,4). Find the coordinates of the fourth vertex. [Hint: Diagonals of a parallelogram have the same mid-point.]

Why the midpoint condition works

Solution

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Q.Three consecutive vertices of a parallelogram ABCD are $A(6,-2,4)$ , $B(2,4,-8)$ , $C(-2,2,4)$ . Find the coordinates of the fourth vertex. [Hint: Diagonals of a parallelogram have the same mid-point.]

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