Collinearity of points MCQ Quiz - Objective Question with Answer for Collinearity of points - Download Free PDF

Given:

Points: (-5, 1), (1, k), and (4, -2)

Formula used:

For three points to be collinear, the area of the triangle formed by them must be zero.

Area of a triangle using coordinates:

Area = (1/2) × [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Calculation:

Since the points are collinear:

(1/2) × [-5(k + 2) + 1(-2 - 1) + 4(1 - k)] = 0

-5(k + 2) + 1(-3) + 4(1 - k) = 0

-5k - 10 - 3 + 4 - 4k = 0

-9k - 9 = 0

-9k = 9

k = -1

∴ The value of k is -1.

Calculation

Since a line passes through P, Q and R, PQ must be a linear multiple of PR.

i.e., $2\hat{i}-6\hat{j}+(2\lambda -2)\hat{k}$ = n($-\hat{i}+3\hat{j}+2\hat{k}$)

Equating LHS and RHS, we get $n=-2$ and $\lambda =-1$

Hence option 3 is correct

Concept -

If point (α1, β1, γ_1), (α2, β2, γ2), (α3, β3, γ2) are collinear, then $\frac{{\alpha }_1-{\alpha }_2}{{\alpha }_2-{\alpha }_3}=\frac{{\beta }_1-{\beta }_2}{{\beta }_2-{\beta }_3}=\frac{{\gamma }_1-{\gamma }_2}{{\gamma }_2-{\gamma }_3}$. 

Explanation -

Given: Points with position vectors

$\alpha \hat{i}+10\hat{j}+13\hat{k},6\hat{i}+11\hat{j}+11\hat{k}$

and $\frac{9}{2}\hat{i}+\beta \hat{j}-8\hat{k}$ are collinear.

So,  $\frac{\alpha -6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta }=\frac{13-11}{11+8}$

⇒ $\frac{2(\alpha -6)}{3}=\frac{-1}{11-\beta }=\frac{2}{19}$

⇒ $\frac{2}{3}(\alpha -6)=\frac{2}{19}$ 

⇒ 19α - 114 = 3 ⇒ 19α = 117 

⇒ $\alpha =\frac{117}{19}$

And, $\frac{-1}{11-\beta }=\frac{2}{19}$

⇒ -19 = 22 - 2β

⇒ 2β = 41

⇒ $\beta =\frac{41}{2}$

∴ $(19\alpha -6\beta {)}^2={(19\times \frac{117}{19}-\frac{6\times 41}{2})}^2$ 

= (117 - 123)² = 36

Hence Option (2) is correct.

Given: 

Collinear points (a,0), (0,b) and (1,1)

Concept:

If points are collinear then

Area of the triangle having coordinates as (x₁ , y₁), (x₂ , y₂), and (x₃ , y₃) is 0

i.e., $\frac{1}{2}$|x1(y2 - y3) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| = 0

Solution:

⇒  $\frac{1}{2}$|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = 0
⇒  |a(b - 1) + 0(1 - 0) + 1(0 - b)| = 0

⇒  ab - a - b = 0

⇒ ab = a + b

⇒ 1 = $\frac{a+b}{ab}$

⇒ $\frac{1}{a}+\frac{1}{b}=1$

⇒ a^-1 + b^-1 = 1

Hence, Option 3 is correct

Concept:

1). The area of the triangle formed by three vertices that are collinear is zero.

2). Area of a triangle formed by (x1​,y1​), (x2​,y2​), and (x3​,y3​) is given by,

     Area = ${\displaystyle \frac{1}{2}}$​[x1​(y2 ​- y3​) + x2​(y3 ​− y1​) + x3​(y1 ​− y2​)]

Calculation:

Given,

Points with coordinates (-5, 0), (5p2, 10p) and (5q2, 10q) are collinear

∴  Area of △ABC=0

We know that,

Area of a triangle formed by (x₁​,y₁​), (x₂​,y₂​), and (x₃​,y₃​) is given by,

Area = ${\displaystyle \frac{1}{2}}$​[x₁​(y₂ ​- y₃​) + x₂​(y₃ ​− y₁​) + x₃​(y₁ ​− y₂​)]

∴ Area of △ABC = ${\displaystyle \frac{1}{2}}$[-5(10p − 10q) + 5p²(10q − 0) + 5q²(0 − 10p)]

⇒ 0 = -50p + 50q + 50p2q - 50q2p

⇒ 0 = 50(q - p) + 50(p2q - q2p)

⇒ 50(p - q) = 50(p2q - q2p)

(p - q) = (p2q - q2p)

(p - q)  = pq(p - q) 

⇒ pq = 1           [p ≠ q]

∴ The value of pq = 1

CONCEPT:

If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then $\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=0$

CONCEPT:

If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then $\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=0$

Concept:

If three points are collinear then the area of the triangle formed from these 3 points will be 0.

[as the collinear points lie on the same straight line]

Analysis:

Area of triangle passing through the points

(x, y), (x₂, y₂) and (x₃, y₃) is given by:

$A=\frac{1}{2}\left[x,\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

(x₁, y₁ = α, β); (x₂, y₂ = 5, 0); (x₃, y₃ = 0, 5)

∴ $A=\frac{1}{2}\left[\alpha \left(-5\right)+5\left(5-\beta \right)+0\right]$ 

-5α + 25 – 5β = 0

α + β = 5

Concept:

If three points (x1, y1), (x2, y2) and (x3, y3) are collinear then the area of the triangle determined by the three points is zero.

$\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|=0$

or

If three or more points are collinear then the slope of any two pairs of points is same.

For example, let three points A, B and C are collinear then

slope of AB = slope of BC = slope of AC

Slope of the line if two points $({\mathrm{x}}_1,{\mathrm{y}}_1)\mathrm{a}\mathrm{n}\mathrm{d}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ are given by: 

$\Rightarrow \left(\mathbf{m}\right)=\frac{{\mathbf{y}}_2-{\mathbf{y}}_1}{{\mathbf{x}}_2-{\mathbf{x}}_1}$

Calculation:

Given: the points (a, 0), (0, b) and (1, 1)  are collinear

$$\begin{gathered} \left|\begin{array}{ccc}a& 0& 1\\ 0& b& 1\\ 1& 1& 1\end{array}\right|=0 \\ \Rightarrow a(b-1)-(0)(0-1)+1(0-b)=0 \end{gathered}$$

ab - a - b = 0

a + b = ab

$\frac{1}{a}+\frac{1}{b}=1$

Concept:

For the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) to be collinear

$\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=0$

Calculation:

The given points are (1, 3, -2), (2, λ, 1) and (3, 4, -1)

The equation of the plane will be 

$\left|\begin{array}{ccc}1& 3& -2\\ 2& \lambda & 1\\ 3& 4& -1\end{array}\right|=0$

⇒ 1(-λ - 4) - 3(-2 - 3) + (-2)(8 - 3λ) = 0

⇒ 5λ - 4 + 15 - 16 = 0

⇒ 5λ = 5

⇒ λ = 1

Formula used:

Area of a triangle:

Let coordinates be A(x1, y1), B(x2, y2) and C(x3, y3),

then the area of a triangle is given as,

Δ =  $\frac{1}{2}$[x1 ( y2 -  y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 )]

Note: If the point A, B, C are collinear then 

The area of Δ ABC should be equal to zero.

Calculation:

Given that,

(-1, 3), (2, p) and (5, -1) are collinear.

⇒ Area of Δ ABC = 0

∵ Δ =  $\frac{1}{2}$[x1 ( y2 -  y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 )]

⇒ $\frac{1}{2}$[-1{p - (-1)} + 2(-1 - 3) + {5(3 - p)] = 0

⇒ -p - 1 + 2 × (-4) + 15 - 5p = 0

⇒ -6p + 6 = 0

⇒ p = 1

Concept:

1). The area of the triangle formed by three vertices that are collinear is zero.

2). Area of a triangle formed by (x1​,y1​), (x2​,y2​), and (x3​,y3​) is given by,

     Area = ${\displaystyle \frac{1}{2}}$​[x1​(y2 ​- y3​) + x2​(y3 ​− y1​) + x3​(y1 ​− y2​)]

Calculation:

Given,

Points with coordinates (-5, 0), (5p2, 10p) and (5q2, 10q) are collinear

∴  Area of △ABC=0

We know that,

Area of a triangle formed by (x₁​,y₁​), (x₂​,y₂​), and (x₃​,y₃​) is given by,

Area = ${\displaystyle \frac{1}{2}}$​[x₁​(y₂ ​- y₃​) + x₂​(y₃ ​− y₁​) + x₃​(y₁ ​− y₂​)]

∴ Area of △ABC = ${\displaystyle \frac{1}{2}}$[-5(10p − 10q) + 5p²(10q − 0) + 5q²(0 − 10p)]

⇒ 0 = -50p + 50q + 50p2q - 50q2p

⇒ 0 = 50(q - p) + 50(p2q - q2p)

⇒ 50(p - q) = 50(p2q - q2p)

(p - q) = (p2q - q2p)

(p - q)  = pq(p - q) 

⇒ pq = 1           [p ≠ q]

∴ The value of pq = 1

Concept:

Let A(x₁, y_1, z₁) and B(x₂, y₂, z₂)

• $\mathrm{D}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{A}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{B}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2+{\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)}^2}$

Calculation:

Given points are P(3, 2, 4), Q(4, 5, 2), R(5, 8, 0) and S(2 -1, 6).

$\mathrm{P}\mathrm{Q}=\sqrt{{\left(4-3\right)}^2+{\left(5-2\right)}^2+{\left(2-4\right)}^2}=\sqrt{1+9+4}=\sqrt{14}$

$\mathrm{Q}\mathrm{R}=\sqrt{{\left(5-4\right)}^2+{\left(8-5\right)}^2+{\left(0-2\right)}^2}=\sqrt{1+9+4}=\sqrt{14}$ 

$\mathrm{R}\mathrm{S}=\sqrt{{\left(2-5\right)}^2+{\left(-1-8\right)}^2+{\left(6-0\right)}^2}=\sqrt{9+81+36}=\sqrt{126}=3\sqrt{14}$

$\mathrm{P}\mathrm{S}=\sqrt{{\left(2-3\right)}^2+{\left(-1-2\right)}^2+{\left(6-4\right)}^2}=\sqrt{1+9+4}=\sqrt{14}$ 

Here, We can write 3√14 = √14 + √14 + √14

⇒ RS = PQ + QR + PS

So, points are collinear.

Given:

(x₁, y₁) = (a, 0)

(x₂, y₂) = (0, b)

(x₃, y₃) = (1, 1)

Formula Used:

Area of triangle = (1/2)[x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

Concept:

If points are collinear, then the area of triangle is zero.

Calculation: 

Area of triangle = (1/2)[x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)] = 0

⇒ Area of triangle = (1/2)[a (b - 1) + 0 (1 - 0) + 1 (0 - b)] = 0

⇒ Area of triangle = (1/2)[ab - a - b] = 0

⇒ Area of triangle = ab - a - b = 0

⇒ ab = a + b

$\Rightarrow \frac{a+b}{ab}$ = 1

Hence, $\frac{a+b}{ab}=1$

Formula used:

Area of triangle:

Let coordinates be A(x1, y1), B(x2, y2) and C(x3, y3),

then the area of a triangle is given as,

Δ =  $\frac{1}{2}$[x1 ( y2 -  y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 )]

Calculation:

Given that,

A(1, 6), B(3, -4) and C(x, y) are collinear.

⇒ Area of Δ ABC = 0

∵ Δ =  $\frac{1}{2}$[x1 ( y2 -  y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 )]

⇒ $\frac{1}{2}$[1(-4 - y) + 3(y - 6) + x[6 - (-4)] = 0

⇒ -4 - y + 3y - 18 + 6x + 4x = 0

⇒ 10x + 2y - 22 = 0

⇒ 5x + y - 11 = 0