Centroid MCQ Quiz in मराठी - Objective Question with Answer for Centroid - मोफत PDF डाउनलोड करा

Calculation

Centroid G divides MR in 1 ∶ 2   

By using section formula 

⇒ G(1, 2, 2) 

\(\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}=k\)

⇒ x = 2

\(\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}=m\)

⇒ x = 1 + m ⇒ 2 = 1 + m ⇒ m = 1

⇒ y = -6, z = 0

Point of intersection A of given lines is (2,–6, 0) 

\(\mathrm{AG}=\sqrt{69}\)

Hence option 3 is correct

Calculation

Given 

L: \(\rm \frac{x+3}{8}=\frac{y-4}{2}=\frac{x+1}{2}=k\)

Point on line L = P(8k - 3, 2k + 4, 2k - 1)

Distance of P from R(1,2,3) is 6 units

⇒ \((8k−4)^2+(2k+2)^2+(2k−4)^2=36\)

⇒ \((64k^2−64k+16)+(4k^2+8k+4)+(4k^2−16k+16)=36\)

⇒ 72k² - 72k + 36 = 36

⇒ 72k2 - 72k = 0

⇒ k = 0,1

At k = 0 ⇒ P(-3,4,-1)

At k = 1 ⇒ Q(5,6,1)

The centroid of the triangle PQR is 

⇒ \((\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3})\)

⇒ (1,4,1) = (α, β, γ)

α2 + β2 + γ2 = 1 + 16 + 1 = 18

Hence option 3 is correct

Given :

(a cos θ1, a sin θ1), (a cos θ2, a sin θ2) and (a cos θ3, a sin θ3) represents the vertices of an equilateral triangle which is inscribed in a circle.

Formula used :

Centroid (x, y) = \((\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2+ y_3}{3})\)         ------ (1)

Calculations :

We know, equilateral triangle is inscribed in a circle with centre (0, 0)

We know from the above figure that

M is the circumcentre of the equilateral \(\triangle ABC\)

M is also the centroid of \(\triangle ABC\)

Using equation (1), we get

⇒ \(\text M = (\frac {a\ cos \theta_1+a\ cos \theta_2+a\ cos \theta_3}{3}, \frac {a\ sin \theta_1+a\ sin \theta_2+a\ sin \theta_3}{3}) = (0, 0)\)

⇒ \(a\ cos \theta_1+a\ cos \theta_2+a\ cos \theta_3 = 0\)

⇒ \(a\ sin \theta_1+a\ sin \theta_2+a\ sin \theta_3 = 0\)

∴ Both statements I and II are correct.

CONCEPT:

If the coordinates of the vertices of triangle Δ ABC are: A (x1, y1), B (x2, y2) and C (x3, y3). Then the co-ordinates of the centroid G is given by: \(\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)

CALCULATION:

Given: (1, 2), (h, - 3) and (- 4, k) are the vertices of triangle and  the centroid of the triangle is located at (5, - 1).

Let A = (1, 2), B = (h, - 3) and C = (- 4, k) are the vertices of Δ ABC.

As we know that, the centroid of a triangle is given by: \(\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)

Here, x₁ = 1, y₁ = 2, x₂ = h, y₂ = - 3, x₃ = - 4 and y₃ = k

⇒ \(G = \left( {\frac{{{1} + {h} - {4}}}{3},\frac{{{2} - {3} + {k}}}{3}} \right)\)

So, the cenroid of the given triangle is: \((\frac{h - 3}{3}, \frac{k - 1}{3})\)

∵ It is given that the centroid of the triangle is located at (5, - 1).

⇒ \(\frac{h - 3}{3} = 5 \ and\ \frac{k - 1}{3} = - 1\)

⇒ h = 18 and k = - 2

⇒ h + k = 16

Hence, option A is the correct answer.

Concept:

Centroid of a Triangle: The point where the three medians of the triangle intersect.

Let’s consider a triangle. If the three vertices of the triangle are A(x₁, y₁), B(x₂, y₂), C(x₃, y₃).

Centroid of a triangle = \(\left( {\frac{{{{\rm{x}}_1} + {\rm{\;}}{{\rm{x}}_2} + {\rm{\;}}{{\rm{x}}_3}}}{3},\frac{{{{\rm{y}}_1} + {\rm{\;}}{{\rm{y}}_2} + {\rm{\;}}{{\rm{y}}_3}}}{3}} \right)\)

Calculation:

Vertices of the triangle are given as (7, x), (y, -6) and (9, 10)

Centroid of a triangle = (6, 3)

⇒ Centroid of a triangle = (6, 3) = \(\left( {\frac{{7 + {\rm{\;y}} + {\rm{\;}}9}}{3},\frac{{{\rm{x}} - 6 + {\rm{\;}}10}}{3}} \right)\)

⇒ (18, 9) = [(16 + y), (x + 4)]

∴ 16 + y = 18 and x + 4 = 9

⇒ y = 2 and x = 5

So, (x, y) = (5, 2)

Concept:

To reflect a point  \((x_1, y_1) \)  across the line ax + by + c = 0, the reflected point (x', y') is given by:

\(x' = x_1 - \frac{2a(ax_1 + by_1 + c)}{a^2 + b^2}, \quad y' = y_1 - \frac{2b(ax_1 + by_1 + c)}{a^2 + b^2}. \)   

Explanation:  

Vertex 1: (1, 3)

\(x' = 1 - \frac{2(1(1) + 2(3) - 2)}{1^2 + 2^2}, \quad y' = 3 - \frac{2(2(1) + 4(3) - 4)}{1^2 + 2^2}\\ x' = 1 - \frac{2(1 + 6 - 2)}{5}, \quad y' = 3 - \frac{2(2 + 12 - 4)}{5}\\ x' = 1 - \frac{10}{5}, \quad y' = 3 - \frac{20}{5}\\ x' = -1, \quad y' = -1\\ \)  

Reflected point: (-1, -1) 

Vertex 2: (3, 1)

\(x' = 3 - \frac{2(1(3) + 2(1) - 2)}{1^2 + 2^2}, \quad y' = 1 - \frac{2(2(3) + 4(1) - 4)}{1^2 + 2^2} \\ x' = 3 - \frac{2(3 + 2 - 2)}{5}, \quad y' = 1 - \frac{2(6 + 4 - 4)}{5} \\ x' = 3 - \frac{6}{5}, \quad y' = 1 - \frac{12}{5} \\ x' = \frac{15}{5} - \frac{6}{5}, \quad y' = \frac{5}{5} - \frac{12}{5} \\ x' = \frac{9}{5}, \quad y' = -\frac{7}{5} \\ \)   

Reflected point:   \(\left(\frac{9}{5}, -\frac{7}{5}\right) \)
Vertex 3: (2, 4)

\(x' = 2 - \frac{2(1(2) + 2(4) - 2)}{1^2 + 2^2}, \quad y' = 4 - \frac{2(2(2) + 4(4) - 4)}{1^2 + 2^2} \\ x' = 2 - \frac{2(2 + 8 - 2)}{5}, \quad y' = 4 - \frac{2(4 + 16 - 4)}{5} \\ x' = 2 - \frac{16}{5}, \quad y' = 4 - \frac{32}{5} \\ x' = \frac{10}{5} - \frac{16}{5}, \quad y' = \frac{20}{5} - \frac{32}{5} \\ x' = -\frac{6}{5}, \quad y' = -\frac{12}{5} \\ \)

Reflected point:  \( \left(-\frac{6}{5}, -\frac{12}{5}\right) \)

Centroid of  \(\Delta PQR \) :

\(\alpha = \frac{x_1 + x_2 + x_3}{3}, \quad \beta = \frac{y_1 + y_2 + y_3}{3} \)   

Substitute the reflected points:

\(\alpha = \frac{-1 + \frac{9}{5} - \frac{6}{5}}{3}, \quad \beta = \frac{-1 - \frac{7}{5} - \frac{12}{5}}{3} \) 


\(\alpha = \frac{-\frac{5}{5} + \frac{9}{5} - \frac{6}{5}}{3}, \quad \beta = \frac{-\frac{5}{5} - \frac{7}{5} - \frac{12}{5}}{3} \)  

\(\alpha = \frac{-\frac{2}{5}}{3}, \quad \beta = \frac{-\frac{24}{5}}{3} \)   


\(\alpha = -\frac{2}{15}, \quad \beta = -\frac{24}{15} \)   

\(15(\alpha - \beta) = 15\left(-\frac{2}{15} - \left(-\frac{24}{15}\right)\right) \)   

\(15(\alpha - \beta) = 15\left(-\frac{2}{15} + \frac{24}{15}\right) \)  

\(15(\alpha - \beta) = 15\left(\frac{22}{15}\right) \)  

\(15(\alpha - \beta) = 22 \)

Hence 22 is the correct answer.

Calculation:

Let the coordinates of C be C = (x, y, z)

Given, A(3, 1, 4) and B(-4, 5, −3) and the centroid of the triangle is G(-1, 2, 1)

∴ \(\frac{3+(-4)+x}{3}\) = - 1

⇒ x = - 2

Also, \(\frac{1+5+y}{3}\) = 2

⇒ y = 0

and, \(\frac{4+(-3)+z}{3}\) = 1

⇒ z = 2

∴ C = (x, y, z) = (- 2, 0, 2)

∴ The coordinates of C are (-2, 0, 2)

The correct answer is Option 2.

Calculation

ΔABC is equilateral

Orthocentre and centroid will be same

⇒ \(\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)\)

Mid - point of AB is D \(\left(\frac{3}{2}, 2, \frac{3}{2}\right)\)

∴ ℓ₁ = \(\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}}\)

⇒ ℓ1 = \(\sqrt{\frac{1}{6}}=\ell_2=\ell_3\)

∴ \(\ell_1^2+\ell_2^2+\ell_3^2=\frac{1}{2}\) 

Hence option (2} is correct

CONCEPT:

If the coordinates of the vertices of triangle Δ ABC are: A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃). Then the co-ordinates of the centroid G is given by: \(\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)

CALCULATION:

Given: A(4,4), B(-1,-2) and C(6,-8) are the coordinates of the triangle ABC.

As we know that, centroid of a triangle ABC whose vertices are A (x1, y1), B (x2, y2) and C (x3, y3) is given by: \(\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)

Here, x₁ = 4, y₁ = 4, x₂ = - 1, y₂ = 2, x₃ = 6 and y₃ = 8.

So, the centroid of triangle ABC is: \(\left( {\frac{{{4} - {1} + {6}}}{3},\frac{{{4} - {2} - {8}}}{3}} \right)\)

=  \(\left(3, - 2 \right)\)

So, the centroid of the given triangle ABC is (3, - 2)

Important Points

The Centroid of a triangle separates the medians in the ratio of 2 : 1

The Centroid of a triangle: The centroid of a triangle is formed when three medians of a triangle intersect. It is one of the four points of concurrency of a triangle. The medians of a triangle are constructed when the vertices of a triangle are joined with the midpoint of the opposite sides of the triangle.