Collinearity of Four Points

Collinearity in geometry refers to the property of points lying on a single straight line. When we discuss the collinearity of four points, we are interested in determining whether all four points reside on the same line. This concept is essential in vector algebra and analytical geometry.

Understanding Collinearity

To understand collinearity, we must first understand the concept of vectors and how they can represent points in space.

Vectors and Points

A vector is a quantity that has both magnitude and direction. In the context of points in space, we can think of a vector as an arrow pointing from the origin of a coordinate system to a point ( P(x, y, z) ). The vector representing this point is often written as ( \vec{OP} ), where ( O ) is the origin.

Conditions for Collinearity

For two points, they are trivially collinear since a line can always be drawn through any two points. However, for three or more points, the condition for collinearity is that the vectors representing the points must be scalar multiples of each other.

For Three Points

Let's say we have three points ( A, B, ) and ( C ). The vectors representing these points are ( \vec{OA}, \vec{OB}, ) and ( \vec{OC} ). The points ( A, B, ) and ( C ) are collinear if and only if there exists a scalar ( k ) such that:

[ \vec{OB} = k \vec{OA} ]

and

[ \vec{OC} = l \vec{OA} ]

where ( l ) is another scalar.

For Four Points

For four points ( A, B, C, ) and ( D ), with vectors ( \vec{OA}, \vec{OB}, \vec{OC}, ) and ( \vec{OD} ), the points are collinear if there exist scalars ( k, l, ) and ( m ) such that:

[ \vec{OB} = k \vec{OA} ] [ \vec{OC} = l \vec{OA} ] [ \vec{OD} = m \vec{OA} ]

Determining Collinearity

To determine if four points are collinear, we can use the following methods:

Method 1: Vector Equations

We can express the vectors ( \vec{AB}, \vec{AC}, ) and ( \vec{AD} ) in terms of the position vectors of the points. If ( \vec{AB}, \vec{AC}, ) and ( \vec{AD} ) are all scalar multiples of each other, then the points are collinear.

Method 2: Cross Product

For points in three-dimensional space, we can use the cross product to determine collinearity. The cross product of two vectors is a vector that is perpendicular to the plane containing the original vectors. If the cross product of ( \vec{AB} ) and ( \vec{AC} ) is zero, then ( A, B, ) and ( C ) are collinear. We can then check if ( \vec{AD} ) is also a scalar multiple of ( \vec{AB} ) or ( \vec{AC} ).

Method 3: Slope

In two-dimensional space, we can use the concept of the slope. If the slope of the line through points ( A ) and ( B ) is the same as the slope of the line through points ( A ) and ( C ), and the slope of the line through points ( A ) and ( D ), then the points are collinear.

Table of Differences and Important Points

Property	Two Points	Three Points	Four Points
Definition	Always collinear	Collinear if vectors are scalar multiples	Collinear if all vectors are scalar multiples
Vector Equation	Not applicable	( \vec{OB} = k \vec{OA} )	( \vec{OB} = k \vec{OA} ), ( \vec{OC} = l \vec{OA} ), ( \vec{OD} = m \vec{OA} )
Cross Product	Not applicable	( \vec{AB} \times \vec{AC} = \vec{0} )	( \vec{AB} \times \vec{AC} = \vec{0} ) and ( \vec{AD} ) is a scalar multiple
Slope (2D)	Not applicable	Same slope for ( AB ) and ( AC )	Same slope for ( AB ), ( AC ), and ( AD )

Examples

Example 1: Vector Equations

Let's say we have four points ( A(1, 2, 3), B(2, 4, 6), C(3, 6, 9), ) and ( D(4, 8, 12) ). The vectors are:

[ \vec{OA} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}, \vec{OB} = \begin{bmatrix} 2 \ 4 \ 6 \end{bmatrix}, \vec{OC} = \begin{bmatrix} 3 \ 6 \ 9 \end{bmatrix}, \vec{OD} = \begin{bmatrix} 4 \ 8 \ 12 \end{bmatrix} ]

It's clear that ( \vec{OB} = 2\vec{OA} ), ( \vec{OC} = 3\vec{OA} ), and ( \vec{OD} = 4\vec{OA} ). Therefore, points ( A, B, C, ) and ( D ) are collinear.

Example 2: Cross Product

Consider points ( A(1, 0, 0), B(2, 0, 0), C(3, 0, 0), ) and ( D(5, 0, 0) ). The vectors ( \vec{AB} ) and ( \vec{AC} ) are:

[ \vec{AB} = \vec{OB} - \vec{OA} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}, \vec{AC} = \vec{OC} - \vec{OA} = \begin{bmatrix} 2 \ 0 \ 0 \end{bmatrix} ]

The cross product ( \vec{AB} \times \vec{AC} ) is:

[ \vec{AB} \times \vec{AC} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} ]

Since the cross product is zero, ( A, B, ) and ( C ) are collinear. Checking ( \vec{AD} ), we find that it is also a scalar multiple of ( \vec{AB} ), confirming that all four points are collinear.

Example 3: Slope (2D)

For points ( A(1, 1), B(2, 2), C(3, 3), ) and ( D(4, 4) ) in two-dimensional space, the slopes are:

[ m_{AB} = \frac{2 - 1}{2 - 1} = 1, m_{AC} = \frac{3 - 1}{3 - 1} = 1, m_{AD} = \frac{4 - 1}{4 - 1} = 1 ]

Since all slopes are equal, the points ( A, B, C, ) and ( D ) are collinear.

In conclusion, determining the collinearity of four points involves checking if the vectors representing the points are scalar multiples of each other, using cross products in three-dimensional space, or comparing slopes in two-dimensional space.