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

Given:

Points are (5, 2, 4), (6, -1, 2) & (8, -7, k)

Concept used:

Area of the triangle formed by using these points = 0

$\frac{1}{2}$ ×= 0

Calculation:

⇒ 5(- k + 14) - 2(6k - 16) + 4(- 42 + 8) = 0

⇒ - 5k + 70 - 12k + 32 + 4(- 34) = 0

⇒ -17k + 102 - 136 = 0

⇒ -17k - 34 = 0

⇒ -17k = 34

⇒ k = -2

Hence the correct answer is "-2".

Concept:

Three points are collinear if the area formed by the three points is 0.

Calculation:

Let the given points be A(1, 1) = (x₁, y₁)

B(–1,–1) = (x₂, y₂)

C$(\sqrt{3},\sqrt{3})$ = (x₃, y₃)

∴ Area of ΔABC = $\frac{1}{2}$|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

= $\frac{1}{2}$|1(-1 - $\sqrt{3}$) - ($\sqrt{3}$ - 1) + $\sqrt{3}$(1 - (-1)|

= $\frac{1}{2}$|- 1 - $\sqrt{3}$ - $\sqrt{3}$ + 1 + 2$\sqrt{3}$|

= ​0​

⇒ The points are collinear

∴ (1, 1), (–1,–1) and $(\sqrt{3},\sqrt{3})$ are collinear points.

The correct answer is Option 2.

Given:

Points are (5, 2, 4), (6, -1, 2) & (8, -7, k)

Concept used:

Area of the triangle formed by using these points = 0

$\frac{1}{2}$ ×= 0

Calculation:

⇒ 5(- k + 14) - 2(6k - 16) + 4(- 42 + 8) = 0

⇒ - 5k + 70 - 12k + 32 + 4(- 34) = 0

⇒ -17k + 102 - 136 = 0

⇒ -17k - 34 = 0

⇒ -17k = 34

⇒ k = -2

Hence the correct answer is "-2".

Given:

Points are (5, 2, 4), (6, -1, 2) & (8, -7, k)

Concept used:

Area of the triangle formed by using these points = 0

$\frac{1}{2}$ ×= 0

Calculation:

⇒ 5(- k + 14) - 2(6k - 16) + 4(- 42 + 8) = 0

⇒ - 5k + 70 - 12k + 32 + 4(- 34) = 0

⇒ -17k + 102 - 136 = 0

⇒ -17k - 34 = 0

⇒ -17k = 34

⇒ k = -2

Hence the correct answer is "-2".

Concept:

Three points are collinear if the area formed by the three points is 0.

Calculation:

Let the given points be A(1, 1) = (x₁, y₁)

B(–1,–1) = (x₂, y₂)

C$(\sqrt{3},\sqrt{3})$ = (x₃, y₃)

∴ Area of ΔABC = $\frac{1}{2}$|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

= $\frac{1}{2}$|1(-1 - $\sqrt{3}$) - ($\sqrt{3}$ - 1) + $\sqrt{3}$(1 - (-1)|

= $\frac{1}{2}$|- 1 - $\sqrt{3}$ - $\sqrt{3}$ + 1 + 2$\sqrt{3}$|

= ​0​

⇒ The points are collinear

∴ (1, 1), (–1,–1) and $(\sqrt{3},\sqrt{3})$ are collinear points.

The correct answer is Option 2.

Concept:

If the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area (A) of ΔABC is given as; 

$A=\frac{1}{2}\cdot \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Calculation: 

Here, we have to find the value of a for  which the points (5, -2), (8, -3) and (a, -12) are collinear

Let,

x1 = 5, y1 = -2,

x2 = 8, y2 = -3,

x3 = a, y3 = -12.

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a Δ ABC then area of Δ ABC =   $A=\frac{1}{2}\cdot \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$\Rightarrow A=\frac{1}{2}\cdot \left|\begin{array}{ccc}5& -2& 1\\ 8& -3& 1\\ a& -12& 1\end{array}\right|$

2A = 5 (-3 + 12) + 2(8 - a) + 1(-96 + 3a)

2A = 45 + 16 - 2a - 96 + 3a

2A = a - 35  

⇒ A = (a - 35)/2

∵ The given points are collinear.

As we know that, if the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

⇒ (a - 35)/2 = 0

⇒ a = 35

Hence, option D is the correct answer.

Alternate Method 

Concept:

Three or more points are collinear if the slope of any two pairs of points is the same.

Calculation:

Let, A = (5, -2), B = (8, -3), C = (a, -12)

Now, the slope of AB = Slope of BC = Slope of AC      (∵ points are collinear)

 $$\begin{gathered} \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{\mathrm{a}-8} \\ \Rightarrow \frac{-1}{3}=\frac{-9}{\mathrm{a}-8} \end{gathered}$$

⇒ a - 8= 27

⇒ a = 27 + 8 = 35

Hence, option (4) is correct.

Concept:

If the point A = (x1, y1), B = (x2, y2) and C = (x3, y3) are collinear then , then the area of triangle ABC is zero.

The area of triangle ABC with vertices A = (x1, y1), B = (x2, y2) and C = (x3, y3) is given by, 

Area = ${\displaystyle \frac{1}{2}}[{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)]$

Calculations:

Given points A = (k, 2k) = (x1, y1), B = (3k, 3k) = (x2, y2) and C = (3, 1) = (x3, y3) are collinear

If the point A = (x1, y1), B = (x2, y2) and C = (x3, y3) are collinear then , then the area of triangle ABC is zero.

The area of triangle ABC with vertices A = (x1, y1), B = (x2, y2) and C = (x3, y3) is given by, 

Area = ${\displaystyle \frac{1}{2}}[{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)]$

⇒ 0 = ${\displaystyle \frac{1}{2}}[\mathrm{k}(3\mathrm{k}-1)+3\mathrm{k}(1-2\mathrm{k})+3(2\mathrm{k}-3\mathrm{k})]$

⇒ 0 = $[3{\mathrm{k}}^2-\mathrm{k}+3\mathrm{k}-6{\mathrm{k}}^2-3\mathrm{k})]$

⇒ 0 = $[-3{\mathrm{k}}^2-\mathrm{k}]$

⇒ k = $-{\displaystyle \frac{1}{3}}$

Hence, the points (k, 2k); (3k, 3k) and (3, 1) are collinear, then the vale of k is = $-{\displaystyle \frac{1}{3}}$

Concept:

1. 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.

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

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

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

slope of AB = slope of BC = slope of AC

The 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 that,

(x, -1), (2, 1) and (4, 5) are collinear

⇒ $\left|\begin{array}{ccc}\mathrm{x}& -1& 1\\ 2& 1& 1\\ 4& 5& 1\end{array}\right|=0$

⇒ x[1×1 - 1×5] + 1[2×1 - 4×1] + 1[2×5 - 1×4] = 0

⇒ -4x - 2 + 6 = 0

⇒ -4x = -4

⇒ x = 1

Concept:

If the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area (A) of ΔABC is given as; 

$A=\frac{1}{2}\cdot \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Calculation: 

Here, we have to find the value of a for  which the points (5, -2), (8, -3) and (a, -12) are collinear

Let,

x1 = 5, y1 = -2,

x2 = 8, y2 = -3,

x3 = a, y3 = -12.

As we know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a Δ ABC then area of Δ ABC =   $A=\frac{1}{2}\cdot \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$\Rightarrow A=\frac{1}{2}\cdot \left|\begin{array}{ccc}5& -2& 1\\ 8& -3& 1\\ a& -12& 1\end{array}\right|$

2A = 5 (-3 + 12) + 2(8 - a) + 1(-96 + 3a)

2A = 45 + 16 - 2a - 96 + 3a

2A = a - 35  

⇒ A = (a - 35)/2

∵ The given points are collinear.

As we know that, if the points A (x1, y1), B (x2, y2) and C (x3, y3) are collinear then area of ΔABC = 0.

⇒ (a - 35)/2 = 0

⇒ a = 35

Hence, option D is the correct answer.

Alternate Method 

Concept:

Three or more points are collinear if the slope of any two pairs of points is the same.

Calculation:

Let, A = (5, -2), B = (8, -3), C = (a, -12)

Now, the slope of AB = Slope of BC = Slope of AC      (∵ points are collinear)

 $$\begin{gathered} \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{\mathrm{a}-8} \\ \Rightarrow \frac{-1}{3}=\frac{-9}{\mathrm{a}-8} \end{gathered}$$

⇒ a - 8= 27

⇒ a = 27 + 8 = 35

Hence, option (4) is correct.

Concept:

If the point A = (x1, y1), B = (x2, y2) and C = (x3, y3) are collinear then , then the area of triangle ABC is zero.

The area of triangle ABC with vertices A = (x1, y1), B = (x2, y2) and C = (x3, y3) is given by, 

Area = ${\displaystyle \frac{1}{2}}[{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)]$

Calculations:

Given points A = (k, 2k) = (x1, y1), B = (3k, 3k) = (x2, y2) and C = (3, 1) = (x3, y3) are collinear

If the point A = (x1, y1), B = (x2, y2) and C = (x3, y3) are collinear then , then the area of triangle ABC is zero.

The area of triangle ABC with vertices A = (x1, y1), B = (x2, y2) and C = (x3, y3) is given by, 

Area = ${\displaystyle \frac{1}{2}}[{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)]$

⇒ 0 = ${\displaystyle \frac{1}{2}}[\mathrm{k}(3\mathrm{k}-1)+3\mathrm{k}(1-2\mathrm{k})+3(2\mathrm{k}-3\mathrm{k})]$

⇒ 0 = $[3{\mathrm{k}}^2-\mathrm{k}+3\mathrm{k}-6{\mathrm{k}}^2-3\mathrm{k})]$

⇒ 0 = $[-3{\mathrm{k}}^2-\mathrm{k}]$

⇒ k = $-{\displaystyle \frac{1}{3}}$

Hence, the points (k, 2k); (3k, 3k) and (3, 1) are collinear, then the vale of k is = $-{\displaystyle \frac{1}{3}}$

Given:

Points are (5, 2, 4), (6, -1, 2) & (8, -7, k)

Concept used:

Area of the triangle formed by using these points = 0

$\frac{1}{2}$ ×= 0

Calculation:

⇒ 5(- k + 14) - 2(6k - 16) + 4(- 42 + 8) = 0

⇒ - 5k + 70 - 12k + 32 + 4(- 34) = 0

⇒ -17k + 102 - 136 = 0

⇒ -17k - 34 = 0

⇒ -17k = 34

⇒ k = -2

Hence the correct answer is "-2".

Concept:

1. 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.

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

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

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

slope of AB = slope of BC = slope of AC

The 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 that,

(x, -1), (2, 1) and (4, 5) are collinear

⇒ $\left|\begin{array}{ccc}\mathrm{x}& -1& 1\\ 2& 1& 1\\ 4& 5& 1\end{array}\right|=0$

⇒ x[1×1 - 1×5] + 1[2×1 - 4×1] + 1[2×5 - 1×4] = 0

⇒ -4x - 2 + 6 = 0

⇒ -4x = -4

⇒ x = 1

Concept:

Three points are collinear if the area formed by the three points is 0.

Calculation:

Let the given points be A(1, 1) = (x₁, y₁)

B(–1,–1) = (x₂, y₂)

C$(\sqrt{3},\sqrt{3})$ = (x₃, y₃)

∴ Area of ΔABC = $\frac{1}{2}$|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

= $\frac{1}{2}$|1(-1 - $\sqrt{3}$) - ($\sqrt{3}$ - 1) + $\sqrt{3}$(1 - (-1)|

= $\frac{1}{2}$|- 1 - $\sqrt{3}$ - $\sqrt{3}$ + 1 + 2$\sqrt{3}$|

= ​0​

⇒ The points are collinear

∴ (1, 1), (–1,–1) and $(\sqrt{3},\sqrt{3})$ are collinear points.

The correct answer is Option 2.

Given:

Points are (5, 2, 4), (6, -1, 2) & (8, -7, k)

Concept used:

Area of the triangle formed by using these points = 0

$\frac{1}{2}$ ×= 0

Calculation:

⇒ 5(- k + 14) - 2(6k - 16) + 4(- 42 + 8) = 0

⇒ - 5k + 70 - 12k + 32 + 4(- 34) = 0

⇒ -17k + 102 - 136 = 0

⇒ -17k - 34 = 0

⇒ -17k = 34

⇒ k = -2

Hence the correct answer is "-2".

Given:

Points are (5, 2, 4), (6, -1, 2) & (8, -7, k)

Concept used:

Area of the triangle formed by using these points = 0

$\frac{1}{2}$ ×= 0

Calculation:

⇒ 5(- k + 14) - 2(6k - 16) + 4(- 42 + 8) = 0

⇒ - 5k + 70 - 12k + 32 + 4(- 34) = 0

⇒ -17k + 102 - 136 = 0

⇒ -17k - 34 = 0

⇒ -17k = 34

⇒ k = -2

Hence the correct answer is "-2".