Show that the points A(a, b+c), B(b, c+a) and C(c, – Class 12 Mathematics | ConceptFirst

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A(a, b+c), B(b, c+a)

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Concept understanding — Collinearity Condition

Collinearity Condition

Three points are collinear when they lie on one straight line. The question this concept answers is: given points $A$ , $B$ , $C$ , how do we test — using vectors, without drawing — whether they fall on a single line?

The Idea

If $A$ , $B$ , $C$ lie on one line, then travelling from $A$ to $B$ and from $B$ to $C$ means moving along the same direction. So the vector $\vec{AB}$ must be a scalar multiple of $\vec{BC}$ : the two segments are parallel and share the point $B$ , which forces all three points onto one line.

A,\,B,\,C

Both forms say the same thing: parallel direction vectors sharing a common point. The cross-product form is convenient because two parallel vectors always have zero cross product.

Using Position Vectors

If $A$ , $B$ , $C$ have position vectors $\vec{a}$ , $\vec{b}$ , $\vec{c}$ , then $\vec{AB}=\vec{b}-\vec{a}$ and $\vec{BC}=\vec{c}-\vec{b}$ , so the test becomes

(\vec{b}-\vec{a})\times(\vec{c}-\vec{b})=\vec{0}.

A Quick Example

Take $A(1,2,3)$ , $B(2,4,5)$ , $C(4,8,9)$ :

$\vec{AB}=(1,2,2)$
$\vec{BC}=(2,4,4)=2\,(1,2,2)=2\,\vec{AB}$

Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$ , the three points are collinear.

Note

In 2D there is an equivalent area test: $A$ , $B$ , $C$ are collinear exactly when the area of triangle $ABC$ is $0$ , i.e. $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0.$ …

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Three points are collinear if the determinant giving the area of the triangle they form equals $0$ .

Area $=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

With $A(a,b+c)$ , $B(b,c+a)$ , $C(c,a+b)$ :

$=\dfrac12\big|a[(c+a)-(a+b)]+b[(a+b)-(b+c)]+c[(b+c)-(c+a)]\big|$

$=\dfrac12\big|a(c-b)+b(a-c)+c(b-a)\big|$

$=\dfrac12\big|ac-ab+ab-bc+bc-ac\big|=\dfrac12|0|=0$

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$m$
›Reveal solutionSolution
Three points are collinear if the vectors between them are parallel (scalar multiples). Using the condition that the cross product of two such vectors is zero, we find $m = 8$ , which corresponds to option (A).

The key idea: collinearity of three points means they lie on a single straight line. In vector terms, if we take any two vectors formed by these points (say from the first to the second, and from the first to the third), they must be parallel — one is a scalar multiple of the other. This gives us a clean algebraic condition.

Let’s label the points:
$A = -\hat{i} - \hat{j} + 2\hat{k}, \quad B = 2\hat{i} + m\hat{j} + 5\hat{k}, \quad C = 3\hat{i} + 11\hat{j} + 6\hat{k}.$

Form two vectors from a common point. Choose $A$ as the reference. Then:

$\vec{AB} = B - A = (2 - (-1))\hat{i} + (m - (-1))\hat{j} + (5 - 2)\hat{k} = 3\hat{i} + (m+1)\hat{j} + 3\hat{k}.$ $\vec{AC} = C - A = (3 - (-1))\hat{i} + (11 - (-1))\hat{j} + (6 - 2)\hat{k} = 4\hat{i} + 12\hat{j} + 4\hat{k}.$

Apply the collinearity condition. For $A$ , $B$ , $C$ to be collinear, $\vec{AB}$ and $\vec{AC}$ must be parallel. That means there exists a scalar $\lambda$ such that:

$\vec{AB} = \lambda \, \vec{AC}.$
Equating components:
$3 = \lambda \cdot 4, \quad m+1 = \lambda \cdot 12, \quad 3 = \lambda \cdot 4.$

Solve for $\lambda$ from the first (or third) equation. From $3 = 4\lambda$ , we get:

$\lambda = \frac{3}{4}.$

Use $\lambda$ to find $m$ . Substitute into the second equation:

$m + 1 = \frac{3}{4} \times 12 = 9.$
Hence:

$m = 9 - 1 = 8.$ …
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CBSE 2026Set ANNUAL1 markMCQ
$A(1,3)$
(a) $\begin{vmatrix}1 & 3 & 1\\ 0 & 0 & 1\\ x & y & 1\end{vmatrix}=0$
(b) $\begin{vmatrix}1 & -3 & 1\\ 0 & 0 & 1\\ x & y & 1\end{vmatrix}=0$
(c) $\begin{vmatrix}1 & 2 & 1\\ 0 & 0 & 1\\ x & y & 1\end{vmatrix}=0$
(d) $\begin{vmatrix}1 & -2 & 1\\ 0 & 0 & 1\\ x & y & 1\end{vmatrix}=0$
›Reveal solutionSolution

The equation of the line through two points $(x_1,y_1)$ and $(x_2,y_2)$ can be written as the determinant condition $\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x&y&1\end{vmatrix}=0$ .

Formula: Three points $(x_1,y_1),(x_2,y_2),(x,y)$ are collinear if and only if

$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x&y&1\end{vmatrix}=0$

Here $A(1,3)$ and $B(0,0)$ are the two fixed points, and $(x,y)$ is a general point on the line. Substituting $(x_1,y_1)=(1,3)$ and $(x_2,y_2)=(0,0)$ :

$\begin{vmatrix}1&3&1\\0&0&1\\x&y&1\end{vmatrix}=0$ …

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CBSE 2025Set ANNUAL1 mark
$(1,2)$
›Reveal solutionSolution

Three collinear points give a zero determinant; expand it to get the line's equation.

A point $(x,y)$ lies on the line through $(1,2)$ and $(3,6)$ iff the three points are collinear:

$\begin{vmatrix}x&y&1\\1&2&1\\3&6&1\end{vmatrix}=0.$

Expanding along the first row:

$x(2\cdot1-1\cdot6)-y(1\cdot1-1\cdot3)+1(1\cdot6-2\cdot3)=0$

$x(2-6)-y(1-3)+(6-6)=0$

$-4x+2y+0=0\ \Rightarrow\ 2y=4x\ \Rightarrow\ y=2x.$

…

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CBSE 20241 markMCQ
$Q. If the points A(3, - 2), B(k, 2) and C(8, 8) are collinear, then the value of k is : (A) 2 (B) - 3 (C) 5 (D) -$
›Reveal solutionSolution
Collinear points lie on the same straight line, so the area of the triangle formed by them is zero. Using the determinant formula for area, we set it to zero and solve for $k$ , obtaining $k = 5$ .

Concept and Intuition: The Collinearity Condition

Three points are collinear if they lie on a single straight line. A powerful geometric fact is that if three points are collinear, the triangle they form has zero area. This gives us a clean algebraic condition: the area of triangle ABC, computed using coordinates, must equal zero.

The standard formula for the area of a triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ is:
$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$
For collinearity, we set this area to zero. Since the absolute value is zero only when the expression inside is zero, we can drop the absolute value and the factor $\frac{1}{2}$ , and simply require:
$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$
This is the collinearity condition — a direct, exam-friendly tool.
$A(x_1, y_1)$
Now, let's apply it step by step.

Assign the coordinates clearly.

We have $A(3, -2)$ , $B(k, 2)$ , $C(8, 8)$ .

So: $x_1 = 3$ , $y_1 = -2$ ; $x_2 = k$ , $y_2 = 2$ ; $x_3 = 8$ , $y_3 = 8$ .

Write the collinearity condition.

Plug into the formula:

$3(2 - 8) + k(8 - (-2)) + 8((-2) - 2) = 0$

Simplify each term carefully.

First term: $3(2 - 8) = 3 \times (-6) = -18$

Second term: $k(8 + 2) = k \times 10 = 10k$

Third term: $8((-2) - 2) = 8 \times (-4) = -32$

So the equation becomes:

$-18 + 10k - 32 = 0$

Combine constants and solve for $k$ . $-18 - 32 = -50$ , so: …
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CBSE 2024Set ANNUAL1 markMCQ
$A(-2\hat{i}+3\hat{j}+5\hat{k})$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Check collinearity via proportional direction ratios of $\overline{AB}$ and $\overline{BC}$ ; check R via the magnitude sum.

Given $A(-2,3,5)$ , $B(1,2,3)$ , $C(7,0,-3)$ (reading $C = 7\hat i + 0\hat j - 3\hat k$ as printed).

$\overline{AB} = B-A = (1-(-2),\,2-3,\,3-5) = (3,-1,-2)$

$\overline{BC} = C-B = (7-1,\,0-2,\,-3-3) = (6,-2,-6)$

$\overline{AC} = C-A = (7-(-2),\,0-3,\,-3-5) = (9,-3,-8)$

Testing collinearity: Three points are collinear iff $\overline{AB}$ and $\overline{BC}$ are parallel, i.e. their components are in the same ratio. Comparing $(3,-1,-2)$ and $(6,-2,-6)$ :

$\frac{6}{3}=2,\quad \frac{-2}{-1}=2,\quad \frac{-6}{-2}=3$

The ratios are $2,2,3$ — not all equal — so $\overline{AB}$ and $\overline{BC}$ are NOT parallel, and hence $A$ , $B$ , $C$ are not collinear. Assertion (A) is FALSE.

Testing Reason (R): $|\overline{AB}|=\sqrt{9+1+4}=\sqrt{14}\approx 3.742$ , $|\overline{BC}|=\sqrt{36+4+36}=\sqrt{76}\approx 8.718$ , $|\overline{AC}|=\sqrt{81+9+64}=\sqrt{154}\approx 12.410$ . …

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CBSE 2024Set ANNUAL1 mark
$(1, 2)$
›Reveal solutionSolution

The line through two given points can be written as a determinant equation set to zero; expand it to get the line's equation.

The line joining $(x_1,y_1)=(1,2)$ and $(x_2,y_2)=(3,6)$ through a general point $(x,y)$ satisfies:

$\begin{vmatrix}x & y & 1\\ 1 & 2 & 1\\ 3 & 6 & 1\end{vmatrix} = 0$

Expanding along the first row:

…

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CBSE 2024Set ANNUAL1 mark
$Q. Using determinants, show that the points (1, 3), (2, 2) and (0, 4) are collinear.$
›Reveal solutionSolution

Three points are collinear if and only if the determinant formed with their coordinates (and a column of 1's) is zero.

Three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are collinear iff

$\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix} = 0$

Here $(x_1,y_1)=(1,3)$ , $(x_2,y_2)=(2,2)$ , $(x_3,y_3)=(0,4)$ .

$\begin{vmatrix}1 & 3 & 1\\ 2 & 2 & 1\\ 0 & 4 & 1\end{vmatrix}$

Expanding along the first row:

$= 1(2\cdot1-1\cdot4) - 3(2\cdot1-1\cdot0) + 1(2\cdot4-2\cdot0)$ …

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CBSE 2023Set ANNUAL1 markMCQ
$x$
(a) 8
(b) 9
(c) 10
(d) 11
›Reveal solutionSolution

Equate the slope of AB with the slope of BC (collinear points share one slope).

Slope of AB $=\dfrac{-1-1}{1-5}=\dfrac{-2}{-4}=\dfrac12$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$\frac{x-2}{a} = \frac{y-3}{b} = \frac{z-4}{c}$
(a) $4a + 2b + 3c = 0$
(b) $4a = 2b = 3c$
(c) $\frac{a}{4} = \frac{b}{2} = \frac{c}{3}$
(d) None of these
›Reveal solutionSolution

Two lines are parallel iff their direction ratios are proportional.

The first line has direction ratios $(a,b,c)$ ; the second has $(4,2,3)$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$x = 0$
(a) $x = -5$
(b) $y = 0$
(c) $z = 5$
(d) None of these
›Reveal solutionSolution

Planes parallel to $x=0$ have the form $x=$ constant; $x=-5$ qualifies.

The plane $x=0$ (the yz-plane) has normal $(1,0,0)$ . A parallel plane shares this normal, so it has the form $x=k$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$2x - 3y + 4z = 7$
(a) $2x - 3y - 4z = 7$
(b) $2x - 3y + 4z = 11$
(c) $2x + 4y - 3z = 11$
(d) None of these
›Reveal solutionSolution

Parallel planes have identical coefficients of $x,y,z$ ; only the constant differs.

…

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CBSE 2021Set NC1 mark
$K$
›Reveal solutionSolution

Three points are collinear iff the determinant formed from their coordinates (with a column of 1's) is zero; expand and solve for $K$ .

Points $A(3,-2)$ , $B(K,2)$ , $C(8,8)$ are collinear iff

$\begin{vmatrix} 3 & -2 & 1\\ K & 2 & 1 \\ 8 & 8 & 1\end{vmatrix}=0$

Expand along the first row:

$3\begin{vmatrix}2&1\\8&1\end{vmatrix} - (-2)\begin{vmatrix}K&1\\8&1\end{vmatrix} + 1\begin{vmatrix}K&2\\8&8\end{vmatrix}=0$

$3(2\cdot1-1\cdot8) + 2(K\cdot1-1\cdot8) + 1(8K-16) = 0$

$3(-6) + 2(K-8) + 8K-16 = 0$

$-18+2K-16+8K-16=0$

$10K - 50 = 0$

$K = 5$

Check (slope method): slope of $AC=\dfrac{8-(-2)}{8-3}=2$ ; slope of $AB=\dfrac{2-(-2)}{5-3}=\dfrac{4}{2}=2$ -- equal, confirming collinearity.

…

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