88283
The coordinates of the point on the line through the points \(A(3,4,1)\) and \(B(5,1,6)\) crosses XY-plane are
1 \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
2 \(\left(\frac{3}{5}, \frac{2}{5}, 0\right)\)
3 \((1,1,0)\)
4 \(\left(-\frac{13}{5}, \frac{23}{5}, 0\right)\)
Explanation:
(A) : Given, The co-ordinate of the points \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) Now, The equation of line in space can be written as \(\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{y}_{2}-\mathrm{y}_{1}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{z}_{2}-\mathrm{z}_{1}} \quad\left[\begin{array}{l}\text { Where }\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right) \text { and } \\ \left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right) \text { are the points }\end{array}\right]\) \(\frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}\) \(\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-4}{-3}=\frac{\mathrm{z}-1}{5}=\mathrm{k}\) Now, In xy plane \(\mathrm{z}=0\) \(\mathrm{k}=-\frac{1}{5}\) \(\therefore \frac{x-3}{2}=\frac{-1}{5} \text { and } \frac{y-4}{-3}=-\frac{1}{5}\) \(x-3=-\frac{2}{5} \text { and } y-4=3 / 5\) \(x=-\frac{2}{5}+3 , y=\frac{3}{5}+4\) \(x=\frac{13}{5} \text { and } y=\frac{23}{5}\) Hence, the required co-ordinate \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
[JCECE-2018]
Co-Ordinate system
88284
If \(y=f(x)\) makes positive intercepts of 2 and 1 unit an \(x\) and \(y\) coordinate axes and encloses an area of \(\frac{3}{4}\) square unit with the axes, then \(\int_{0}^{2} x f^{\prime}(x) d x\) is is
1 \(\frac{3}{2}\)
2 1
3 \(\frac{5}{4}\)
4 \(-\frac{3}{4}\)
Explanation:
(D) : Given, \(\int_{0}^{2} f(x) d x=3 / 4\) \(\therefore \quad \int_{0}^{2} x f^{\prime}(x) d x\) \(=x \int_{0}^{2} f^{\prime}(x) d x-\int_{0}^{2} f(x) d x\) \(=\left[\mathrm{x} \int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right]_{0}^{2}-3 / 4=2 \mathrm{f}(2)-3 / 4\) \(=2 \times 0-3 / 4\) \([\mathrm{y}=\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{y}=\mathrm{f}(2)]=-3 / 4\)
BCECE-2018
Co-Ordinate system
88285
If \(A(2,3)\) and \(B(-2,1)\) are two vertices of a triangle and third vertex moves on the line \(2 x+3 y=9\), then the locus of the centroid of the triangle is
1 \(2 x+3 y=1\)
2 \(2 x+y=3\)
3 \(2 x-3 y=1\)
4 \(x-y=1\)
Explanation:
(A) : Given, The co-ordinate of vertices of triangle are \(\mathrm{A}(2,-3), \mathrm{B}(-\) \(2,1)\) and \(\mathrm{C}(\mathrm{x}, \mathrm{y})\) Now, Let the co-ordinate of centroid be \((h, k)\) \(\therefore \quad \mathrm{h}=\left(\frac{2-2+\mathrm{x}}{3}\right) \quad \Rightarrow \quad \mathrm{x}=3 \mathrm{~h} \) \(\mathrm{k}=\left(\frac{-3+1+\mathrm{y}}{3}\right) \Rightarrow \mathrm{y}=3 \mathrm{k}+2\) Where \((x, y)\) are the point on the line \(2 x+3 y=9\) \(\therefore \quad 2(3 h)+3(3 k+2)=9\) \(6 \mathrm{~h}+9 \mathrm{k}+6=9\) \(6 \mathrm{~h}+9 \mathrm{k}=3\) \(2 \mathrm{~h}+3 \mathrm{k}=1\) Locus of the centroid of the triangle is \(2 x+3 y=1\)
BCECE-2017
Co-Ordinate system
88287
Let \(A(2,-3)\) and \(B(-2,1)\) be vertices of a \(\triangle \mathrm{ABC}\). If The centroid of this triangle moves on the line \(2 x+3 y=1\), then the locus of the vertex \(C\) is the line
1 \(2 x+3 y=9\)
2 \(2 x-3 y=7\)
3 \(3 x+2 y=5\)
4 \(3 x-2 y=3\)
Explanation:
(A) : Given, The co-ordinate of vertices of triangle \(\mathrm{A}(2,-3), \mathrm{B}(-2,1)\) and \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) Now, co-ordinate of centroid lies on the line \(2 x+3 y=1\) \(\Rightarrow \quad\left(\frac{2-2+\mathrm{h}}{3}, \frac{-3+1+\mathrm{k}}{3}\right) \Rightarrow\left(\frac{\mathrm{h}}{3}, \frac{-2+\mathrm{k}}{3}\right)\) \(\therefore\left(\frac{\mathrm{h}}{3}, \frac{-2+\mathrm{k}}{3}\right)\) satisfy the equation \(\{2 \mathrm{x}+3 \mathrm{y}=1\}\) \(\Rightarrow 2\left(\frac{\mathrm{h}}{3}\right)+3\left(\frac{-2+\mathrm{k}}{3}\right)=1\) \(\frac{2}{3} \mathrm{~h}+(-2+\mathrm{k})=1\) \(\Rightarrow \quad 2 \mathrm{~h}-6+3 \mathrm{k}=3\) \(2 \mathrm{~h}+3 \mathrm{k}=9\) Locus of the vertex \(C\) is the line \(2 x+3 y=9\)
88283
The coordinates of the point on the line through the points \(A(3,4,1)\) and \(B(5,1,6)\) crosses XY-plane are
1 \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
2 \(\left(\frac{3}{5}, \frac{2}{5}, 0\right)\)
3 \((1,1,0)\)
4 \(\left(-\frac{13}{5}, \frac{23}{5}, 0\right)\)
Explanation:
(A) : Given, The co-ordinate of the points \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) Now, The equation of line in space can be written as \(\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{y}_{2}-\mathrm{y}_{1}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{z}_{2}-\mathrm{z}_{1}} \quad\left[\begin{array}{l}\text { Where }\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right) \text { and } \\ \left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right) \text { are the points }\end{array}\right]\) \(\frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}\) \(\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-4}{-3}=\frac{\mathrm{z}-1}{5}=\mathrm{k}\) Now, In xy plane \(\mathrm{z}=0\) \(\mathrm{k}=-\frac{1}{5}\) \(\therefore \frac{x-3}{2}=\frac{-1}{5} \text { and } \frac{y-4}{-3}=-\frac{1}{5}\) \(x-3=-\frac{2}{5} \text { and } y-4=3 / 5\) \(x=-\frac{2}{5}+3 , y=\frac{3}{5}+4\) \(x=\frac{13}{5} \text { and } y=\frac{23}{5}\) Hence, the required co-ordinate \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
[JCECE-2018]
Co-Ordinate system
88284
If \(y=f(x)\) makes positive intercepts of 2 and 1 unit an \(x\) and \(y\) coordinate axes and encloses an area of \(\frac{3}{4}\) square unit with the axes, then \(\int_{0}^{2} x f^{\prime}(x) d x\) is is
1 \(\frac{3}{2}\)
2 1
3 \(\frac{5}{4}\)
4 \(-\frac{3}{4}\)
Explanation:
(D) : Given, \(\int_{0}^{2} f(x) d x=3 / 4\) \(\therefore \quad \int_{0}^{2} x f^{\prime}(x) d x\) \(=x \int_{0}^{2} f^{\prime}(x) d x-\int_{0}^{2} f(x) d x\) \(=\left[\mathrm{x} \int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right]_{0}^{2}-3 / 4=2 \mathrm{f}(2)-3 / 4\) \(=2 \times 0-3 / 4\) \([\mathrm{y}=\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{y}=\mathrm{f}(2)]=-3 / 4\)
BCECE-2018
Co-Ordinate system
88285
If \(A(2,3)\) and \(B(-2,1)\) are two vertices of a triangle and third vertex moves on the line \(2 x+3 y=9\), then the locus of the centroid of the triangle is
1 \(2 x+3 y=1\)
2 \(2 x+y=3\)
3 \(2 x-3 y=1\)
4 \(x-y=1\)
Explanation:
(A) : Given, The co-ordinate of vertices of triangle are \(\mathrm{A}(2,-3), \mathrm{B}(-\) \(2,1)\) and \(\mathrm{C}(\mathrm{x}, \mathrm{y})\) Now, Let the co-ordinate of centroid be \((h, k)\) \(\therefore \quad \mathrm{h}=\left(\frac{2-2+\mathrm{x}}{3}\right) \quad \Rightarrow \quad \mathrm{x}=3 \mathrm{~h} \) \(\mathrm{k}=\left(\frac{-3+1+\mathrm{y}}{3}\right) \Rightarrow \mathrm{y}=3 \mathrm{k}+2\) Where \((x, y)\) are the point on the line \(2 x+3 y=9\) \(\therefore \quad 2(3 h)+3(3 k+2)=9\) \(6 \mathrm{~h}+9 \mathrm{k}+6=9\) \(6 \mathrm{~h}+9 \mathrm{k}=3\) \(2 \mathrm{~h}+3 \mathrm{k}=1\) Locus of the centroid of the triangle is \(2 x+3 y=1\)
BCECE-2017
Co-Ordinate system
88287
Let \(A(2,-3)\) and \(B(-2,1)\) be vertices of a \(\triangle \mathrm{ABC}\). If The centroid of this triangle moves on the line \(2 x+3 y=1\), then the locus of the vertex \(C\) is the line
1 \(2 x+3 y=9\)
2 \(2 x-3 y=7\)
3 \(3 x+2 y=5\)
4 \(3 x-2 y=3\)
Explanation:
(A) : Given, The co-ordinate of vertices of triangle \(\mathrm{A}(2,-3), \mathrm{B}(-2,1)\) and \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) Now, co-ordinate of centroid lies on the line \(2 x+3 y=1\) \(\Rightarrow \quad\left(\frac{2-2+\mathrm{h}}{3}, \frac{-3+1+\mathrm{k}}{3}\right) \Rightarrow\left(\frac{\mathrm{h}}{3}, \frac{-2+\mathrm{k}}{3}\right)\) \(\therefore\left(\frac{\mathrm{h}}{3}, \frac{-2+\mathrm{k}}{3}\right)\) satisfy the equation \(\{2 \mathrm{x}+3 \mathrm{y}=1\}\) \(\Rightarrow 2\left(\frac{\mathrm{h}}{3}\right)+3\left(\frac{-2+\mathrm{k}}{3}\right)=1\) \(\frac{2}{3} \mathrm{~h}+(-2+\mathrm{k})=1\) \(\Rightarrow \quad 2 \mathrm{~h}-6+3 \mathrm{k}=3\) \(2 \mathrm{~h}+3 \mathrm{k}=9\) Locus of the vertex \(C\) is the line \(2 x+3 y=9\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88283
The coordinates of the point on the line through the points \(A(3,4,1)\) and \(B(5,1,6)\) crosses XY-plane are
1 \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
2 \(\left(\frac{3}{5}, \frac{2}{5}, 0\right)\)
3 \((1,1,0)\)
4 \(\left(-\frac{13}{5}, \frac{23}{5}, 0\right)\)
Explanation:
(A) : Given, The co-ordinate of the points \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) Now, The equation of line in space can be written as \(\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{y}_{2}-\mathrm{y}_{1}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{z}_{2}-\mathrm{z}_{1}} \quad\left[\begin{array}{l}\text { Where }\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right) \text { and } \\ \left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right) \text { are the points }\end{array}\right]\) \(\frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}\) \(\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-4}{-3}=\frac{\mathrm{z}-1}{5}=\mathrm{k}\) Now, In xy plane \(\mathrm{z}=0\) \(\mathrm{k}=-\frac{1}{5}\) \(\therefore \frac{x-3}{2}=\frac{-1}{5} \text { and } \frac{y-4}{-3}=-\frac{1}{5}\) \(x-3=-\frac{2}{5} \text { and } y-4=3 / 5\) \(x=-\frac{2}{5}+3 , y=\frac{3}{5}+4\) \(x=\frac{13}{5} \text { and } y=\frac{23}{5}\) Hence, the required co-ordinate \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
[JCECE-2018]
Co-Ordinate system
88284
If \(y=f(x)\) makes positive intercepts of 2 and 1 unit an \(x\) and \(y\) coordinate axes and encloses an area of \(\frac{3}{4}\) square unit with the axes, then \(\int_{0}^{2} x f^{\prime}(x) d x\) is is
1 \(\frac{3}{2}\)
2 1
3 \(\frac{5}{4}\)
4 \(-\frac{3}{4}\)
Explanation:
(D) : Given, \(\int_{0}^{2} f(x) d x=3 / 4\) \(\therefore \quad \int_{0}^{2} x f^{\prime}(x) d x\) \(=x \int_{0}^{2} f^{\prime}(x) d x-\int_{0}^{2} f(x) d x\) \(=\left[\mathrm{x} \int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right]_{0}^{2}-3 / 4=2 \mathrm{f}(2)-3 / 4\) \(=2 \times 0-3 / 4\) \([\mathrm{y}=\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{y}=\mathrm{f}(2)]=-3 / 4\)
BCECE-2018
Co-Ordinate system
88285
If \(A(2,3)\) and \(B(-2,1)\) are two vertices of a triangle and third vertex moves on the line \(2 x+3 y=9\), then the locus of the centroid of the triangle is
1 \(2 x+3 y=1\)
2 \(2 x+y=3\)
3 \(2 x-3 y=1\)
4 \(x-y=1\)
Explanation:
(A) : Given, The co-ordinate of vertices of triangle are \(\mathrm{A}(2,-3), \mathrm{B}(-\) \(2,1)\) and \(\mathrm{C}(\mathrm{x}, \mathrm{y})\) Now, Let the co-ordinate of centroid be \((h, k)\) \(\therefore \quad \mathrm{h}=\left(\frac{2-2+\mathrm{x}}{3}\right) \quad \Rightarrow \quad \mathrm{x}=3 \mathrm{~h} \) \(\mathrm{k}=\left(\frac{-3+1+\mathrm{y}}{3}\right) \Rightarrow \mathrm{y}=3 \mathrm{k}+2\) Where \((x, y)\) are the point on the line \(2 x+3 y=9\) \(\therefore \quad 2(3 h)+3(3 k+2)=9\) \(6 \mathrm{~h}+9 \mathrm{k}+6=9\) \(6 \mathrm{~h}+9 \mathrm{k}=3\) \(2 \mathrm{~h}+3 \mathrm{k}=1\) Locus of the centroid of the triangle is \(2 x+3 y=1\)
BCECE-2017
Co-Ordinate system
88287
Let \(A(2,-3)\) and \(B(-2,1)\) be vertices of a \(\triangle \mathrm{ABC}\). If The centroid of this triangle moves on the line \(2 x+3 y=1\), then the locus of the vertex \(C\) is the line
1 \(2 x+3 y=9\)
2 \(2 x-3 y=7\)
3 \(3 x+2 y=5\)
4 \(3 x-2 y=3\)
Explanation:
(A) : Given, The co-ordinate of vertices of triangle \(\mathrm{A}(2,-3), \mathrm{B}(-2,1)\) and \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) Now, co-ordinate of centroid lies on the line \(2 x+3 y=1\) \(\Rightarrow \quad\left(\frac{2-2+\mathrm{h}}{3}, \frac{-3+1+\mathrm{k}}{3}\right) \Rightarrow\left(\frac{\mathrm{h}}{3}, \frac{-2+\mathrm{k}}{3}\right)\) \(\therefore\left(\frac{\mathrm{h}}{3}, \frac{-2+\mathrm{k}}{3}\right)\) satisfy the equation \(\{2 \mathrm{x}+3 \mathrm{y}=1\}\) \(\Rightarrow 2\left(\frac{\mathrm{h}}{3}\right)+3\left(\frac{-2+\mathrm{k}}{3}\right)=1\) \(\frac{2}{3} \mathrm{~h}+(-2+\mathrm{k})=1\) \(\Rightarrow \quad 2 \mathrm{~h}-6+3 \mathrm{k}=3\) \(2 \mathrm{~h}+3 \mathrm{k}=9\) Locus of the vertex \(C\) is the line \(2 x+3 y=9\)
88283
The coordinates of the point on the line through the points \(A(3,4,1)\) and \(B(5,1,6)\) crosses XY-plane are
1 \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
2 \(\left(\frac{3}{5}, \frac{2}{5}, 0\right)\)
3 \((1,1,0)\)
4 \(\left(-\frac{13}{5}, \frac{23}{5}, 0\right)\)
Explanation:
(A) : Given, The co-ordinate of the points \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) Now, The equation of line in space can be written as \(\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{y}_{2}-\mathrm{y}_{1}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{z}_{2}-\mathrm{z}_{1}} \quad\left[\begin{array}{l}\text { Where }\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right) \text { and } \\ \left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right) \text { are the points }\end{array}\right]\) \(\frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}\) \(\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-4}{-3}=\frac{\mathrm{z}-1}{5}=\mathrm{k}\) Now, In xy plane \(\mathrm{z}=0\) \(\mathrm{k}=-\frac{1}{5}\) \(\therefore \frac{x-3}{2}=\frac{-1}{5} \text { and } \frac{y-4}{-3}=-\frac{1}{5}\) \(x-3=-\frac{2}{5} \text { and } y-4=3 / 5\) \(x=-\frac{2}{5}+3 , y=\frac{3}{5}+4\) \(x=\frac{13}{5} \text { and } y=\frac{23}{5}\) Hence, the required co-ordinate \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
[JCECE-2018]
Co-Ordinate system
88284
If \(y=f(x)\) makes positive intercepts of 2 and 1 unit an \(x\) and \(y\) coordinate axes and encloses an area of \(\frac{3}{4}\) square unit with the axes, then \(\int_{0}^{2} x f^{\prime}(x) d x\) is is
1 \(\frac{3}{2}\)
2 1
3 \(\frac{5}{4}\)
4 \(-\frac{3}{4}\)
Explanation:
(D) : Given, \(\int_{0}^{2} f(x) d x=3 / 4\) \(\therefore \quad \int_{0}^{2} x f^{\prime}(x) d x\) \(=x \int_{0}^{2} f^{\prime}(x) d x-\int_{0}^{2} f(x) d x\) \(=\left[\mathrm{x} \int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right]_{0}^{2}-3 / 4=2 \mathrm{f}(2)-3 / 4\) \(=2 \times 0-3 / 4\) \([\mathrm{y}=\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{y}=\mathrm{f}(2)]=-3 / 4\)
BCECE-2018
Co-Ordinate system
88285
If \(A(2,3)\) and \(B(-2,1)\) are two vertices of a triangle and third vertex moves on the line \(2 x+3 y=9\), then the locus of the centroid of the triangle is
1 \(2 x+3 y=1\)
2 \(2 x+y=3\)
3 \(2 x-3 y=1\)
4 \(x-y=1\)
Explanation:
(A) : Given, The co-ordinate of vertices of triangle are \(\mathrm{A}(2,-3), \mathrm{B}(-\) \(2,1)\) and \(\mathrm{C}(\mathrm{x}, \mathrm{y})\) Now, Let the co-ordinate of centroid be \((h, k)\) \(\therefore \quad \mathrm{h}=\left(\frac{2-2+\mathrm{x}}{3}\right) \quad \Rightarrow \quad \mathrm{x}=3 \mathrm{~h} \) \(\mathrm{k}=\left(\frac{-3+1+\mathrm{y}}{3}\right) \Rightarrow \mathrm{y}=3 \mathrm{k}+2\) Where \((x, y)\) are the point on the line \(2 x+3 y=9\) \(\therefore \quad 2(3 h)+3(3 k+2)=9\) \(6 \mathrm{~h}+9 \mathrm{k}+6=9\) \(6 \mathrm{~h}+9 \mathrm{k}=3\) \(2 \mathrm{~h}+3 \mathrm{k}=1\) Locus of the centroid of the triangle is \(2 x+3 y=1\)
BCECE-2017
Co-Ordinate system
88287
Let \(A(2,-3)\) and \(B(-2,1)\) be vertices of a \(\triangle \mathrm{ABC}\). If The centroid of this triangle moves on the line \(2 x+3 y=1\), then the locus of the vertex \(C\) is the line
1 \(2 x+3 y=9\)
2 \(2 x-3 y=7\)
3 \(3 x+2 y=5\)
4 \(3 x-2 y=3\)
Explanation:
(A) : Given, The co-ordinate of vertices of triangle \(\mathrm{A}(2,-3), \mathrm{B}(-2,1)\) and \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) Now, co-ordinate of centroid lies on the line \(2 x+3 y=1\) \(\Rightarrow \quad\left(\frac{2-2+\mathrm{h}}{3}, \frac{-3+1+\mathrm{k}}{3}\right) \Rightarrow\left(\frac{\mathrm{h}}{3}, \frac{-2+\mathrm{k}}{3}\right)\) \(\therefore\left(\frac{\mathrm{h}}{3}, \frac{-2+\mathrm{k}}{3}\right)\) satisfy the equation \(\{2 \mathrm{x}+3 \mathrm{y}=1\}\) \(\Rightarrow 2\left(\frac{\mathrm{h}}{3}\right)+3\left(\frac{-2+\mathrm{k}}{3}\right)=1\) \(\frac{2}{3} \mathrm{~h}+(-2+\mathrm{k})=1\) \(\Rightarrow \quad 2 \mathrm{~h}-6+3 \mathrm{k}=3\) \(2 \mathrm{~h}+3 \mathrm{k}=9\) Locus of the vertex \(C\) is the line \(2 x+3 y=9\)
