QBManager

Co-Ordinate system

88267 \(\mathbf{A}=(\cos \theta, \sin \theta), B=(\sin \theta,-\cos \theta)\) are two points. The locus of the centroid of \(\triangle O A B\), where ' \(O\) ' is the origin is

1 \(\mathrm{x}^{2}+\mathrm{y}^{2}=3\)

2 \(9 x^{2}+9 y^{2}=2\)

3 \(2 x^{2}+2 y^{2}=9\)

4 \(3 x^{2}+3 y^{2}=2\)

Karnataka CET-2013

Co-Ordinate system

88268 The points \((11,9),(2,1)\) and \((2,-1)\) are the midpoints of the sides of the triangle. Then the centroid is

1 \((5,3)\)

2 \((-5,-3)\)

3 \(5,-3)\)

4 \(3,5)\)

Karnataka CET-2012

Co-Ordinate system

88269 The mid points of the sides of triangle are \((1,5\), \(-1),(0,4,-2)\) and \((2,3,4)\), then centroid of the triangle is

1 \((1,4,3)\)

2 \(\left(1,4, \frac{1}{3}\right)\)

3 \((-1,4,3)\)

4 \(\left(\frac{1}{3}, 2,4\right)\)

Explanation:

(B) : Given,
The co-ordinates of mid points of a triangle
\(\mathrm{ABC}\) are \((1,5,-1),(0,4,-2)\) and \((2,3,4)\)

Let us assume the co-ordinate of \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)\) and \(\left(\mathrm{x}_{3}, \mathrm{y}_{3}, \mathrm{z}_{3}\right)\) respectively. Now,
We know that,
\(\therefore \quad\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}\right]=(2,3,4)\)
\(\left.\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}\right]=(1,5,-1)\)
\({\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{3}}{2}\right]=(0,4,-2)}\)
\(\text { Now, } \quad \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}+\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2}+\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2}=1+2+0\)
\(\therefore \quad \frac{2\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}\right)}{2}=3\)
\(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}=3\)
\(\mathrm{x}_{3}=3-\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)\)
\(=3-4\)
\({\left[\because \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}=2\right.}\)
\(\mathrm{x}_{3}=-1\)
\(\frac{x_{2}+x_{3}}{2}=1\)
\(\mathrm{x}_{2}+\mathrm{x}_{3}=2\)
\(x_{2}-1=2 \quad \Rightarrow \quad x_{2}=3\)
\(\mathrm{x}_{1}=0-\mathrm{x}_{3} \quad \Rightarrow \quad \mathrm{x}_{1}=1\)
Similarly, \(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}=3, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}=5, \frac{\mathrm{y}_{3}+\mathrm{y}_{1}}{2}=4\)
\(\therefore \quad 2\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)=24\)
\(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}=12\)
\(\mathrm{y}_{3}=6, \mathrm{y}_{2}=4, \mathrm{y}_{1}=2\)
and, \(\quad \frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}=4, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}=-1, \frac{\mathrm{z}_{3}+\mathrm{z}_{1}}{2}=-2\)
\(\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}=1\)
\(\therefore \quad \mathrm{z}_{1}=3, \quad \mathrm{z}_{2}=5, \quad \mathrm{z}_{3}=-7\)
\(\therefore \quad \mathrm{A}=(1,2,3)\)
\(\mathrm{B}=(3,4,5)\)
\(\mathrm{C}=(-1,6,-7)\)
Now,
Co-ordinate of centroid
\(\mathrm{G} =\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}, \frac{\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}}{3}\right]\)
\(=\left[\frac{1+3+(-1)}{3}, \frac{2+4+6}{3}, \frac{3+5-7}{3}\right]\)
\(=[1,4,1 / 3]\)

Karnataka CET-2021

Co-Ordinate system

88271 If a vertex of triangle is \((3,3)\) and the mid points of two sides through this vertex are \(\left(2, \frac{2}{3}\right)\) and \(\left(4, \frac{3}{2}\right)\), then the centroid of the triangle is given by

1 \((1,3)\)

2 \((3,0)\)

3 \((3,4 / 9)\)

4 \((0,3)\)

COMEDK-2012

Co-Ordinate system

88267 \(\mathbf{A}=(\cos \theta, \sin \theta), B=(\sin \theta,-\cos \theta)\) are two points. The locus of the centroid of \(\triangle O A B\), where ' \(O\) ' is the origin is

1 \(\mathrm{x}^{2}+\mathrm{y}^{2}=3\)

2 \(9 x^{2}+9 y^{2}=2\)

3 \(2 x^{2}+2 y^{2}=9\)

4 \(3 x^{2}+3 y^{2}=2\)

Karnataka CET-2013

Co-Ordinate system

88268 The points \((11,9),(2,1)\) and \((2,-1)\) are the midpoints of the sides of the triangle. Then the centroid is

1 \((5,3)\)

2 \((-5,-3)\)

3 \(5,-3)\)

4 \(3,5)\)

Karnataka CET-2012

Co-Ordinate system

88269 The mid points of the sides of triangle are \((1,5\), \(-1),(0,4,-2)\) and \((2,3,4)\), then centroid of the triangle is

1 \((1,4,3)\)

2 \(\left(1,4, \frac{1}{3}\right)\)

3 \((-1,4,3)\)

4 \(\left(\frac{1}{3}, 2,4\right)\)

Explanation:

(B) : Given,
The co-ordinates of mid points of a triangle
\(\mathrm{ABC}\) are \((1,5,-1),(0,4,-2)\) and \((2,3,4)\)

Let us assume the co-ordinate of \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)\) and \(\left(\mathrm{x}_{3}, \mathrm{y}_{3}, \mathrm{z}_{3}\right)\) respectively. Now,
We know that,
\(\therefore \quad\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}\right]=(2,3,4)\)
\(\left.\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}\right]=(1,5,-1)\)
\({\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{3}}{2}\right]=(0,4,-2)}\)
\(\text { Now, } \quad \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}+\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2}+\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2}=1+2+0\)
\(\therefore \quad \frac{2\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}\right)}{2}=3\)
\(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}=3\)
\(\mathrm{x}_{3}=3-\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)\)
\(=3-4\)
\({\left[\because \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}=2\right.}\)
\(\mathrm{x}_{3}=-1\)
\(\frac{x_{2}+x_{3}}{2}=1\)
\(\mathrm{x}_{2}+\mathrm{x}_{3}=2\)
\(x_{2}-1=2 \quad \Rightarrow \quad x_{2}=3\)
\(\mathrm{x}_{1}=0-\mathrm{x}_{3} \quad \Rightarrow \quad \mathrm{x}_{1}=1\)
Similarly, \(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}=3, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}=5, \frac{\mathrm{y}_{3}+\mathrm{y}_{1}}{2}=4\)
\(\therefore \quad 2\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)=24\)
\(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}=12\)
\(\mathrm{y}_{3}=6, \mathrm{y}_{2}=4, \mathrm{y}_{1}=2\)
and, \(\quad \frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}=4, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}=-1, \frac{\mathrm{z}_{3}+\mathrm{z}_{1}}{2}=-2\)
\(\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}=1\)
\(\therefore \quad \mathrm{z}_{1}=3, \quad \mathrm{z}_{2}=5, \quad \mathrm{z}_{3}=-7\)
\(\therefore \quad \mathrm{A}=(1,2,3)\)
\(\mathrm{B}=(3,4,5)\)
\(\mathrm{C}=(-1,6,-7)\)
Now,
Co-ordinate of centroid
\(\mathrm{G} =\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}, \frac{\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}}{3}\right]\)
\(=\left[\frac{1+3+(-1)}{3}, \frac{2+4+6}{3}, \frac{3+5-7}{3}\right]\)
\(=[1,4,1 / 3]\)

Karnataka CET-2021

Co-Ordinate system

88271 If a vertex of triangle is \((3,3)\) and the mid points of two sides through this vertex are \(\left(2, \frac{2}{3}\right)\) and \(\left(4, \frac{3}{2}\right)\), then the centroid of the triangle is given by

1 \((1,3)\)

2 \((3,0)\)

3 \((3,4 / 9)\)

4 \((0,3)\)

COMEDK-2012

Co-Ordinate system

88267 \(\mathbf{A}=(\cos \theta, \sin \theta), B=(\sin \theta,-\cos \theta)\) are two points. The locus of the centroid of \(\triangle O A B\), where ' \(O\) ' is the origin is

1 \(\mathrm{x}^{2}+\mathrm{y}^{2}=3\)

2 \(9 x^{2}+9 y^{2}=2\)

3 \(2 x^{2}+2 y^{2}=9\)

4 \(3 x^{2}+3 y^{2}=2\)

Karnataka CET-2013

Co-Ordinate system

88268 The points \((11,9),(2,1)\) and \((2,-1)\) are the midpoints of the sides of the triangle. Then the centroid is

1 \((5,3)\)

2 \((-5,-3)\)

3 \(5,-3)\)

4 \(3,5)\)

Karnataka CET-2012

Co-Ordinate system

88269 The mid points of the sides of triangle are \((1,5\), \(-1),(0,4,-2)\) and \((2,3,4)\), then centroid of the triangle is

1 \((1,4,3)\)

2 \(\left(1,4, \frac{1}{3}\right)\)

3 \((-1,4,3)\)

4 \(\left(\frac{1}{3}, 2,4\right)\)

Explanation:

(B) : Given,
The co-ordinates of mid points of a triangle
\(\mathrm{ABC}\) are \((1,5,-1),(0,4,-2)\) and \((2,3,4)\)

Let us assume the co-ordinate of \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)\) and \(\left(\mathrm{x}_{3}, \mathrm{y}_{3}, \mathrm{z}_{3}\right)\) respectively. Now,
We know that,
\(\therefore \quad\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}\right]=(2,3,4)\)
\(\left.\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}\right]=(1,5,-1)\)
\({\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{3}}{2}\right]=(0,4,-2)}\)
\(\text { Now, } \quad \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}+\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2}+\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2}=1+2+0\)
\(\therefore \quad \frac{2\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}\right)}{2}=3\)
\(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}=3\)
\(\mathrm{x}_{3}=3-\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)\)
\(=3-4\)
\({\left[\because \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}=2\right.}\)
\(\mathrm{x}_{3}=-1\)
\(\frac{x_{2}+x_{3}}{2}=1\)
\(\mathrm{x}_{2}+\mathrm{x}_{3}=2\)
\(x_{2}-1=2 \quad \Rightarrow \quad x_{2}=3\)
\(\mathrm{x}_{1}=0-\mathrm{x}_{3} \quad \Rightarrow \quad \mathrm{x}_{1}=1\)
Similarly, \(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}=3, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}=5, \frac{\mathrm{y}_{3}+\mathrm{y}_{1}}{2}=4\)
\(\therefore \quad 2\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)=24\)
\(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}=12\)
\(\mathrm{y}_{3}=6, \mathrm{y}_{2}=4, \mathrm{y}_{1}=2\)
and, \(\quad \frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}=4, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}=-1, \frac{\mathrm{z}_{3}+\mathrm{z}_{1}}{2}=-2\)
\(\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}=1\)
\(\therefore \quad \mathrm{z}_{1}=3, \quad \mathrm{z}_{2}=5, \quad \mathrm{z}_{3}=-7\)
\(\therefore \quad \mathrm{A}=(1,2,3)\)
\(\mathrm{B}=(3,4,5)\)
\(\mathrm{C}=(-1,6,-7)\)
Now,
Co-ordinate of centroid
\(\mathrm{G} =\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}, \frac{\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}}{3}\right]\)
\(=\left[\frac{1+3+(-1)}{3}, \frac{2+4+6}{3}, \frac{3+5-7}{3}\right]\)
\(=[1,4,1 / 3]\)

Karnataka CET-2021

Co-Ordinate system

88271 If a vertex of triangle is \((3,3)\) and the mid points of two sides through this vertex are \(\left(2, \frac{2}{3}\right)\) and \(\left(4, \frac{3}{2}\right)\), then the centroid of the triangle is given by

1 \((1,3)\)

2 \((3,0)\)

3 \((3,4 / 9)\)

4 \((0,3)\)

COMEDK-2012

Co-Ordinate system

88267 \(\mathbf{A}=(\cos \theta, \sin \theta), B=(\sin \theta,-\cos \theta)\) are two points. The locus of the centroid of \(\triangle O A B\), where ' \(O\) ' is the origin is

1 \(\mathrm{x}^{2}+\mathrm{y}^{2}=3\)

2 \(9 x^{2}+9 y^{2}=2\)

3 \(2 x^{2}+2 y^{2}=9\)

4 \(3 x^{2}+3 y^{2}=2\)

Karnataka CET-2013

Co-Ordinate system

88268 The points \((11,9),(2,1)\) and \((2,-1)\) are the midpoints of the sides of the triangle. Then the centroid is

1 \((5,3)\)

2 \((-5,-3)\)

3 \(5,-3)\)

4 \(3,5)\)

Karnataka CET-2012

Co-Ordinate system

88269 The mid points of the sides of triangle are \((1,5\), \(-1),(0,4,-2)\) and \((2,3,4)\), then centroid of the triangle is

1 \((1,4,3)\)

2 \(\left(1,4, \frac{1}{3}\right)\)

3 \((-1,4,3)\)

4 \(\left(\frac{1}{3}, 2,4\right)\)

Explanation:

(B) : Given,
The co-ordinates of mid points of a triangle
\(\mathrm{ABC}\) are \((1,5,-1),(0,4,-2)\) and \((2,3,4)\)

Let us assume the co-ordinate of \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)\) and \(\left(\mathrm{x}_{3}, \mathrm{y}_{3}, \mathrm{z}_{3}\right)\) respectively. Now,
We know that,
\(\therefore \quad\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}\right]=(2,3,4)\)
\(\left.\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}\right]=(1,5,-1)\)
\({\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{3}}{2}\right]=(0,4,-2)}\)
\(\text { Now, } \quad \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}+\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2}+\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2}=1+2+0\)
\(\therefore \quad \frac{2\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}\right)}{2}=3\)
\(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}=3\)
\(\mathrm{x}_{3}=3-\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)\)
\(=3-4\)
\({\left[\because \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}=2\right.}\)
\(\mathrm{x}_{3}=-1\)
\(\frac{x_{2}+x_{3}}{2}=1\)
\(\mathrm{x}_{2}+\mathrm{x}_{3}=2\)
\(x_{2}-1=2 \quad \Rightarrow \quad x_{2}=3\)
\(\mathrm{x}_{1}=0-\mathrm{x}_{3} \quad \Rightarrow \quad \mathrm{x}_{1}=1\)
Similarly, \(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}=3, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}=5, \frac{\mathrm{y}_{3}+\mathrm{y}_{1}}{2}=4\)
\(\therefore \quad 2\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)=24\)
\(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}=12\)
\(\mathrm{y}_{3}=6, \mathrm{y}_{2}=4, \mathrm{y}_{1}=2\)
and, \(\quad \frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}=4, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}=-1, \frac{\mathrm{z}_{3}+\mathrm{z}_{1}}{2}=-2\)
\(\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}=1\)
\(\therefore \quad \mathrm{z}_{1}=3, \quad \mathrm{z}_{2}=5, \quad \mathrm{z}_{3}=-7\)
\(\therefore \quad \mathrm{A}=(1,2,3)\)
\(\mathrm{B}=(3,4,5)\)
\(\mathrm{C}=(-1,6,-7)\)
Now,
Co-ordinate of centroid
\(\mathrm{G} =\left[\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}, \frac{\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}}{3}\right]\)
\(=\left[\frac{1+3+(-1)}{3}, \frac{2+4+6}{3}, \frac{3+5-7}{3}\right]\)
\(=[1,4,1 / 3]\)

Karnataka CET-2021

Co-Ordinate system

88271 If a vertex of triangle is \((3,3)\) and the mid points of two sides through this vertex are \(\left(2, \frac{2}{3}\right)\) and \(\left(4, \frac{3}{2}\right)\), then the centroid of the triangle is given by

1 \((1,3)\)

2 \((3,0)\)

3 \((3,4 / 9)\)

4 \((0,3)\)

COMEDK-2012

Question Bank Series

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