Show that the points $(a + 5,\, a - 4)$, $(a - 2,\, a + 3)$ and $(a,\, a)$ do not lie on a straight line for any value of $a$.

Show that the points (a + 5, a - 4), (a - 2, a + 3 – Class 12 Mathematics

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(a + 5,\, a - 4)

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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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The triangle on the three points has area $\tfrac72$ for every $a$ — a fixed non-zero value — so the points can never lie on one straight line.

The idea

Three points lie on a straight line precisely when the triangle they span is degenerate, i.e. has zero area. So instead of comparing slopes, compute the area and show it is a non-zero constant.

Set up

Let

A=(a+5,\;a-4),\quad B=(a-2,\;a+3),\quad C=(a,\;a).

The area of $\triangle ABC$ is

\text{Area}=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.

Work the steps

Compute the $y$ -differences:

y_2-y_3=(a+3)-a=3,\quad y_3-y_1=a-(a-4)=4,\quad y_1-y_2=(a-4)-(a+3)=-7.

Substitute:

\text{Area}=\frac12\left|(a+5)(3)+(a-2)(4)+a(-7)\right|.

Expand inside the bars:

3a+15+4a-8-7a=(3a+4a-7a)+(15-8)=0\cdot a+7=7.

Hence …

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

AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$Q. Suppose that the three points A, B and C in the plane are such that their x-coordinates as well as y-coordinates are in GP with the same common ratio. Then the points A, B and C (A) constitute a right angled triangle (B) form an isosceles triangle (C) lie on a straight line (D) form an equilateral triangle$
›Reveal solutionSolution
Points whose $x$ - and $y$ -coordinates are both in GP with the same ratio are all scalar multiples of one fixed vector, hence collinear (and pass through the origin).

Concept and Intuition

If $x$ -coordinates are $a, ar, ar^2$ and $y$ -coordinates are $b, br, br^2$ (same ratio $r$ for both), then the three points are $A=(a,b)$ , $B=(ar,br)=r(a,b)$ , $C=(ar^2,br^2)=r^2(a,b)$ . Every point is just a scalar multiple of the vector $(a,b)$ , so geometrically they all lie on the single line through the origin passing through $(a,b)$ .

Step-by-Step Solution

Let the common first terms be $a$ (for $x$ ) and $b$ (for $y$ ), with common ratio $r$ .

$A=(a,b)$ , $B=(ar,br)$ , $C=(ar^2,br^2)$ .

Factor: $B=r\cdot A$ , $C=r^2\cdot A$ (as position vectors). …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$(\alpha\bar{i} + 10\bar{j} + 13\bar{k})$
›Reveal solutionSolution
Using the collinearity condition on three position vectors and solving for $\alpha,\beta$ , we get $(19\alpha-6\beta)^2=36$ .

Concept and Intuition

Three points with position vectors $P_1,P_2,P_3$ are collinear exactly when $P_3-P_1$ is a scalar multiple of $P_2-P_1$ . Matching components gives simultaneous equations for the unknowns, and it's cleanest to first find the scalar $t$ from whichever component is fully known (here, the $z$ -component).

Step-by-Step Solution

Let $P_1=(\alpha,10,13)$ , $P_2=(6,11,11)$ , $P_3=\left(\tfrac92,\beta,-8\right)$ .

$P_2-P_1=(6-\alpha,\,1,\,-2)$ ; $P_3-P_1=\left(\tfrac92-\alpha,\,\beta-10,\,-21\right)$ .

Collinearity: $P_3-P_1=t(P_2-P_1)$ for some scalar $t$ .

$z$ -component: $-21=t(-2)\Rightarrow t=\dfrac{21}{2}$ .

$y$ -component: $\beta-10=t(1)=\dfrac{21}2 \Rightarrow \beta=10+\dfrac{21}2=\dfrac{41}2$ .

$x$ -component: $\tfrac92-\alpha = t(6-\alpha) = \dfrac{21}2(6-\alpha)$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$\overline{OA} = \bar{a}$
›Reveal solutionSolution
This tests recognizing that $x+y=1$ makes $C$ an affine combination of $A$ and $B$ — i.e. $C$ lies on line $AB$ — and that $C$ sits between $A$ and $B$ (inside the triangle's boundary) exactly when both coefficients are positive.

Concept and Intuition

Any point of the form $x\bar a+y\bar b$ with $x+y=1$ lies on the straight line through $A$ and $B$ , because this is precisely the parametric/affine representation of that line ( $x=1-t,\,y=t$ traces all of line $AB$ as $t$ ranges over all reals). The point lies on the segment $AB$ (not the extended line) exactly when both barycentric-style weights $x,y$ are non-negative — and strictly between $A$ and $B$ when both are strictly positive.

Step-by-Step Solution

Since $x+y=1$ , write $y=1-x$ ; then $C=x\bar a+(1-x)\bar b=\bar b+x(\bar a-\bar b)$ , which is exactly the parametrization of line $AB$ .

As $x$ ranges over $(0,1)$ , $C$ traces every point strictly between $A$ and $B$ (i.e. on the open segment $AB$ ).

If $x\le0$ or $x\ge1$ , $C$ falls outside segment $AB$ , either beyond $B$ or beyond $A$ . …
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Q.Show that the points $(a + 5,\, a - 4)$ , $(a - 2,\, a + 3)$ and $(a,\, a)$ do not lie on a straight line for any value of $a$ .

Concept understanding — Collinearity Condition

Collinearity Condition

The Idea

Using Position Vectors

A Quick Example

The idea

Set up

Work the steps

Method: Area-of-Triangle Determinant Test for Collinearity

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Show that the points (a+5, a−4)(a + 5,\, a - 4)(a+5,a−4), (a−2, a+3)(a - 2,\, a + 3)(a−2,a+3) and (a, a)(a,\, a)(a,a) do not lie on a straight line for any value of aaa.

The idea

Set up

Work the steps

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Q.Show that the points $(a + 5,\, a - 4)$ , $(a - 2,\, a + 3)$ and $(a,\, a)$ do not lie on a straight line for any value of $a$ .

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