Show that the point A(1,-1,3), B(2,-4,5) and (5,-13,11) are collinear.

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A(1,-1,3)

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Concept understanding — Collinearity Slope Condition

Collinearity Slope Condition — From Intuition to Precision

Imagine three points scattered on a sheet of paper. You can always draw a triangle through them. But if those three points happen to lie perfectly on a single straight line — like beads on a thread — they are called collinear (from Latin co- meaning "together" and linearis meaning "belonging to a line").

The question is: how do you check, using only coordinates, whether three given points are collinear?

The Intuition

Think about walking from point A to point B, then from point B to point C. If all three lie on the same line, your direction of travel should not change when you turn at B. In other words, the slope of AB must equal the slope of BC.

Slope measures steepness: $\frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$ . If two segments share the same slope and meet at a common point (B), they lie on the same straight line.

Note

This works because a line has a constant slope everywhere. If AB and BC have the same slope, they are parts of the same line — they cannot bend.

The Precise Statement

Let three points be $A(x_1, y_1)$ , $B(x_2, y_2)$ , and $C(x_3, y_3)$ . They are collinear if and only if:

\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_3 - y_2}{x_3 - x_2}

provided that $x_1 \neq x_2$ and $x_2 \neq x_3$ (i.e., no vertical segment).

\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_3 - y_2}{x_3 - x_2}

What About Vertical Lines?

If $x_1 = x_2$ , the slope of AB is undefined (division by zero). But the condition still works: if AB is vertical, then for collinearity, BC must also be vertical — meaning $x_2 = x_3$ . So the condition becomes: either both slopes are equal and defined, or both are undefined (i.e., both segments are vertical).

A Cleaner Algebraic Form

Cross-multiplying the slope equality gives a form that avoids division entirely:

(y_2 - y_1)(x_3 - x_2) = (y_3 - y_2)(x_2 - x_1)

This works for all cases, including vertical lines.

Tip

For quick checks, use the cross-multiplied form — no need to worry about zero denominators.

Example

Check if $A(1, 2)$ , $B(3, 6)$ , $C(5, 10)$ are collinear.

Slope of AB: $\frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$

Slope of BC: $\frac{10 - 6}{5 - 3} = \frac{4}{2} = 2$

Slopes are equal → points are collinear. Indeed, they all lie on $y = 2x$ .

Why This Matters

The slope condition is the simplest test for collinearity in coordinate geometry. It appears in: …

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To show three points are collinear, we can demonstrate that two vectors formed by these points (e.g., $\vec{AB}$ and $\vec{BC}$ ) are parallel and share a common point. We find that $\vec{BC} = 3\vec{AB}$ , confirming they are parallel and thus the points are collinear.

For three points to be collinear, they must lie on the same straight line. In three-dimensional space, a common way to prove collinearity is by using vectors.

Concept: Collinearity using Vectors

If three points $A$ , $B$ , and $C$ are collinear, then the vector $\vec{AB}$ must be parallel to the vector $\vec{BC}$ (or $\vec{AC}$ ).

Two vectors are parallel if one is a scalar multiple of the other. That is, if $\vec{u}$ and $\vec{v}$ are parallel, then $\vec{u} = k\vec{v}$ for some scalar $k$ .

If $\vec{AB} = k \vec{BC}$ (or $\vec{BC} = k \vec{AB}$ ), and they share a common point (in this case, point $B$ ), then the points $A$ , $B$ , and $C$ must lie on the same line.

Let the given points be $A(x_1, y_1, z_1)$ , $B(x_2, y_2, z_2)$ , and $C(x_3, y_3, z_3)$ .

The vector from point $P_1(x_1, y_1, z_1)$ to $P_2(x_2, y_2, z_2)$ is given by $\vec{P_1P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$ .

We will calculate the vectors $\vec{AB}$ and $\vec{BC}$ and then check if one is a scalar multiple of the other.

Identify the coordinates of the points.

We are given the points:

$A = (1, -1, 3)$

$B = (2, -4, 5)$

$C = (5, -13, 11)$
Calculate the vector $\vec{AB}$ .

$\vec{AB} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j} + (z_B - z_A)\hat{k}$

$\vec{AB} = (2 - 1)\hat{i} + (-4 - (-1))\hat{j} + (5 - 3)\hat{k}$

$\vec{AB} = (1)\hat{i} + (-4 + 1)\hat{j} + (2)\hat{k}$

$\vec{AB} = \hat{i} - 3\hat{j} + 2\hat{k}$
Calculate the vector $\vec{BC}$ .

$\vec{BC} = (x_C - x_B)\hat{i} + (y_C - y_B)\hat{j} + (z_C - z_B)\hat{k}$

$\vec{BC} = (5 - 2)\hat{i} + (-13 - (-4))\hat{j} + (11 - 5)\hat{k}$

$\vec{BC} = (3)\hat{i} + (-13 + 4)\hat{j} + (6)\hat{k}$

$\vec{BC} = 3\hat{i} - 9\hat{j} + 6\hat{k}$
Check if $\vec{AB}$ and $\vec{BC}$ are parallel.

For $\vec{AB}$ and $\vec{BC}$ to be parallel, there must exist a scalar $k$ such that $\vec{BC} = k \vec{AB}$ .

Comparing the components:

$3\hat{i} - 9\hat{j} + 6\hat{k} = k (\hat{i} - 3\hat{j} + 2\hat{k})$

Equating the $\hat{i}$ components:

$3 = k \cdot 1 \implies k = 3$

Equating the $\hat{j}$ components:

$-9 = k \cdot (-3) \implies k = \frac{-9}{-3} = 3$ …

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Q.Show that the point A(1,−1,3)A(1,-1,3)A(1,−1,3), B(2,−4,5)B(2,-4,5)B(2,−4,5) and (5,−13,11)(5,-13,11)(5,−13,11) are collinear.

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Q.Show that the point $A(1,-1,3)$ , $B(2,-4,5)$ and $(5,-13,11)$ are collinear.

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