Prove that the points (0,-1,-7), (2,1,-9) and (6,5,-13) are collinear. Find the ratio in which the first point

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(0,-1,-7)

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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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The points are collinear because the vector formed by two points is a scalar multiple of the vector formed by another pair. The first point divides the join of the other two externally in the ratio $\boxed{1:3}$ .

Concept: Collinearity in 3D

Three points $A$ , $B$ , and $C$ are collinear if they lie on the same straight line. In three-dimensional space, a straightforward way to prove collinearity is by using vectors. If points $A$ , $B$ , and $C$ are collinear, then the vector $\vec{AB}$ must be parallel to the vector $\vec{BC}$ (or $\vec{AC}$ ). This means one vector can be expressed as a scalar multiple of the other. Since they share a common point (e.g., $B$ is common to $\vec{AB}$ and $\vec{BC}$ ), their parallelism implies collinearity.

Alternatively, one can show that the sum of the lengths of two smaller segments equals the length of the largest segment (e.g., $AB + BC = AC$ ).

Proving Collinearity

Let the given points be $A(0,-1,-7)$ , $B(2,1,-9)$ , and $C(6,5,-13)$ .

Calculate the position vectors:

The position vectors corresponding to the points are:

$\vec{A} = (0\hat{i} - 1\hat{j} - 7\hat{k})$

$\vec{B} = (2\hat{i} + 1\hat{j} - 9\hat{k})$

$\vec{C} = (6\hat{i} + 5\hat{j} - 13\hat{k})$
Form vectors between the points:

We form two vectors, for example, $\vec{AB}$ and $\vec{BC}$ .

$\vec{AB} = \vec{B} - \vec{A} = (2-0)\hat{i} + (1-(-1))\hat{j} + (-9-(-7))\hat{k}$

$\vec{AB} = 2\hat{i} + 2\hat{j} - 2\hat{k}$

$\vec{BC} = \vec{C} - \vec{B} = (6-2)\hat{i} + (5-1)\hat{j} + (-13-(-9))\hat{k}$

$\vec{BC} = 4\hat{i} + 4\hat{j} - 4\hat{k}$
Check for scalar proportionality:

Observe the relationship between $\vec{AB}$ and $\vec{BC}$ :

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

$\vec{BC} = 2 \cdot \vec{AB}$

Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$ (with scalar $2$ ), the vectors $\vec{AB}$ and $\vec{BC}$ are parallel. As they share a common point $B$ , the points $A$ , $B$ , and $C$ must be collinear.

Tip

Another way to prove collinearity is to calculate the distances $AB$ , $BC$ , and $AC$ . If $AB+BC=AC$ (or any permutation), the points are collinear.

$AB = \sqrt{2^2+2^2+(-2)^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$

$BC = \sqrt{4^2+4^2+(-4)^2} = \sqrt{16+16+16} = \sqrt{48} = 4\sqrt{3}$

$AC = \sqrt{(6-0)^2+(5-(-1))^2+(-13-(-7))^2} = \sqrt{6^2+6^2+(-6)^2} = \sqrt{36+36+36} = \sqrt{108} = 6\sqrt{3}$

Since $AB+BC = 2\sqrt{3} + 4\sqrt{3} = 6\sqrt{3} = AC$ , the points are collinear. This also shows that point $B$ lies between $A$ and $C$ .

Concept: Ratio of Division (Section Formula)

If a point $P(x,y,z)$ divides the line segment joining $Q(x_1,y_1,z_1)$ and $R(x_2,y_2,z_2)$ in the ratio $m:n$ , its coordinates are given by the section formula:

P(x,y,z) = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)

If the ratio $m:n$ is positive, the division is internal. If the ratio is negative (e.g., $m:(-n)$ ), the division is external. An external division means the point $P$ lies on the line containing $QR$ but outside the segment $QR$ .

Finding the Ratio of Division

We need to find the ratio in which the first point $A(0,-1,-7)$ divides the join of the other two points $B(2,1,-9)$ and $C(6,5,-13)$ . Let $A$ divide $BC$ in the ratio $m:n$ .

Apply the section formula for the x-coordinate: Using the x-coordinates of $A$ , $B$ , and $C$ : …

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Q.Prove that the points (0,−1,−7)(0,-1,-7)(0,−1,−7), (2,1,−9)(2,1,-9)(2,1,−9) and (6,5,−13)(6,5,-13)(6,5,−13) are collinear. Find the ratio in which the first point divides the join of the other two.

Concept: Collinearity in 3D

Proving Collinearity

Concept: Ratio of Division (Section Formula)

Finding the Ratio of Division

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Q.Prove that the points $(0,-1,-7)$ , $(2,1,-9)$ and $(6,5,-13)$ are collinear. Find the ratio in which the first point divides the join of the other two.

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