88210
If the axis are rotated through an angle of \(30^{\circ}\) in the clockwise direction, the point \((4,2 \sqrt{3})\) in the new system is
1 \((2,3)\)
2 \((2, \sqrt{3})\)
3 \((\sqrt{3}, 2)\)
4 \((\sqrt{3}, 5)\)
Explanation:
(D) : Given, The coordinate of \((x, y)\) is \((4,2 \sqrt{3})\) \(x=4, y=2 \sqrt{3}\) \(\theta=-30^{\circ} \text { [Clock wise] }\) We know, \(\mathrm{X}=\mathrm{x} \cos \theta+\mathrm{y} \sin \theta\) and, \(\mathrm{Y}=\mathrm{x} \sin \theta+\mathrm{y} \cos \theta\) \(\therefore \quad \mathrm{X}=4 \cos 30^{\circ}-2 \sqrt{3} \sin 30^{\circ}\) \(=4 \times \frac{\sqrt{3}}{2}-2 \sqrt{3} \times \frac{1}{2}=\sqrt{3}\) \(Y=4 \sin 30^{\circ}+2 \sqrt{3} \cos 30^{\circ}\) \(=4 \times \frac{1}{2}+2 \sqrt{3} \times \frac{\sqrt{3}}{2}=2+3 \quad=5\) The given Point in the new system \((\sqrt{3}, 5)\)
JCECE-2015
Co-Ordinate system
88211
If the axis are rotated through an angle \(45^{\circ}\) in the positive direction without changing the origin, then the co-ordinates of the point \((\sqrt{2}, 4)\) in the old system are
1 \((1-2 \sqrt{2}, 1+2 \sqrt{2})\)
2 \((1+2 \sqrt{2}, 1-2 \sqrt{2})\)
3 \((2 \sqrt{2}, \sqrt{2})\)
4 \((\sqrt{2}, 2)\)
Explanation:
(A) : Given, The axis are rotated through an angle \(=45^{\circ}\) The co-ordinate of the point \(=(\sqrt{2}, 4)\) We know that, \(\mathrm{x}^{\prime}=\mathrm{x} \cos \theta-\mathrm{y} \sin \theta\) \(y^{\prime}=x \sin \theta+y \cos \theta\) Now, \(x^{\prime}=x \cos 45^{\circ}-y \sin 45^{\circ}\) \(x^{\prime}=\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}}-\frac{4}{\sqrt{2}}=1-2 \sqrt{2}\) \(y^{\prime}=x \sin \theta+y \cos \theta\) \(y^{\prime}=\sqrt{2} \sin 45^{\circ}+4 \cos 45^{\circ}\) \(y^{\prime}=\frac{\sqrt{2}}{\sqrt{2}}+\frac{4}{\sqrt{2}}\) \(y^{\prime}=1+2 \sqrt{2}\) \(\therefore\) The co-ordinate is \(\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)=(1-2 \sqrt{2}, 1+2 \sqrt{2})\)
AP EAMCET-2002
Co-Ordinate system
88212
If the axis are transformed to the point \((-1,1)\) then the equation \(3 x^{2}+y^{2}+2 x+4 y+15=0\) would transform to
1 \(3 x^{2}+2 y^{2}-4 x+\overline{6 y+23}=0\)
2 \(3 x^{2}+y^{2}-4 x+6 y+21=0\)
3 \(3 x^{2}+y^{2}+4 x-6 y-21=0\)
4 \(3 x^{2}+y^{2}+4 x+6 y+21=0\)
Explanation:
(B): Given, The axis transformed to the point \((-1,1)\) Initial equation \(3 x^{2}+y^{2}+2 x+4 y+15=0\) We know that, The co-ordinate of transformed point \(\mathrm{x}=\mathrm{x}+\mathrm{h}\) \(\mathrm{y}=\mathrm{y}+\mathrm{k}\) \(\therefore \quad \mathrm{x}=\mathrm{x}-1\) Now, The equation transform to \(3(\mathrm{x}-1)^{2}+(\mathrm{y}+1)^{2}+2(\mathrm{x}-1)+4(\mathrm{y}+1)+15=0\) \(3\left(\mathrm{x}^{2}+1-2 \mathrm{x}\right)+\mathrm{y}^{2}+1+2 \mathrm{y}+2 \mathrm{x}-2+4 \mathrm{y}+4+15=0\) \(3 \mathrm{x}^{2}+3-6 \mathrm{x}+\mathrm{y}^{2}+1+2 \mathrm{y}+2 \mathrm{x}-2+4 \mathrm{y}+4+15=0\) \(3 x^{2}-4 x+y^{2}+6 y+21=0\) \(3 x^{2}+y^{2}-4 x+6 y+21=0\)
APEAPCET-2021-20.08.2021
Co-Ordinate system
88213
When the axes are rotated through an angle \(45^{\circ}\). the new coordinates of a point \(P\) are \((1,-1)\). The coordinates of \(P\) in the original system are
1 \((\sqrt{2}, \sqrt{2)}\)
2 \((\sqrt{2}, 0)\)
3 \(0, \sqrt{2})\)
4 \((-\sqrt{2}, 0)\)
Explanation:
(B) : Given, The angle of rotation of axis \(=45^{\circ}\) The new co-ordinates of a point \(\mathrm{P}=(1,-1)\) We know that, \(x=x \cos \theta-y \sin \theta\) \(y=x \sin \theta+y \cos \theta\) Now, \(x=1 \times \cos 45^{\circ}-(-1) \sin 45^{\circ}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\) \(y=1 \times \sin 45^{\circ}+(-1) \cos 45^{\circ}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\) \(\therefore \quad\) Original co-ordinate \(=(\sqrt{2}, 0)\)
88210
If the axis are rotated through an angle of \(30^{\circ}\) in the clockwise direction, the point \((4,2 \sqrt{3})\) in the new system is
1 \((2,3)\)
2 \((2, \sqrt{3})\)
3 \((\sqrt{3}, 2)\)
4 \((\sqrt{3}, 5)\)
Explanation:
(D) : Given, The coordinate of \((x, y)\) is \((4,2 \sqrt{3})\) \(x=4, y=2 \sqrt{3}\) \(\theta=-30^{\circ} \text { [Clock wise] }\) We know, \(\mathrm{X}=\mathrm{x} \cos \theta+\mathrm{y} \sin \theta\) and, \(\mathrm{Y}=\mathrm{x} \sin \theta+\mathrm{y} \cos \theta\) \(\therefore \quad \mathrm{X}=4 \cos 30^{\circ}-2 \sqrt{3} \sin 30^{\circ}\) \(=4 \times \frac{\sqrt{3}}{2}-2 \sqrt{3} \times \frac{1}{2}=\sqrt{3}\) \(Y=4 \sin 30^{\circ}+2 \sqrt{3} \cos 30^{\circ}\) \(=4 \times \frac{1}{2}+2 \sqrt{3} \times \frac{\sqrt{3}}{2}=2+3 \quad=5\) The given Point in the new system \((\sqrt{3}, 5)\)
JCECE-2015
Co-Ordinate system
88211
If the axis are rotated through an angle \(45^{\circ}\) in the positive direction without changing the origin, then the co-ordinates of the point \((\sqrt{2}, 4)\) in the old system are
1 \((1-2 \sqrt{2}, 1+2 \sqrt{2})\)
2 \((1+2 \sqrt{2}, 1-2 \sqrt{2})\)
3 \((2 \sqrt{2}, \sqrt{2})\)
4 \((\sqrt{2}, 2)\)
Explanation:
(A) : Given, The axis are rotated through an angle \(=45^{\circ}\) The co-ordinate of the point \(=(\sqrt{2}, 4)\) We know that, \(\mathrm{x}^{\prime}=\mathrm{x} \cos \theta-\mathrm{y} \sin \theta\) \(y^{\prime}=x \sin \theta+y \cos \theta\) Now, \(x^{\prime}=x \cos 45^{\circ}-y \sin 45^{\circ}\) \(x^{\prime}=\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}}-\frac{4}{\sqrt{2}}=1-2 \sqrt{2}\) \(y^{\prime}=x \sin \theta+y \cos \theta\) \(y^{\prime}=\sqrt{2} \sin 45^{\circ}+4 \cos 45^{\circ}\) \(y^{\prime}=\frac{\sqrt{2}}{\sqrt{2}}+\frac{4}{\sqrt{2}}\) \(y^{\prime}=1+2 \sqrt{2}\) \(\therefore\) The co-ordinate is \(\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)=(1-2 \sqrt{2}, 1+2 \sqrt{2})\)
AP EAMCET-2002
Co-Ordinate system
88212
If the axis are transformed to the point \((-1,1)\) then the equation \(3 x^{2}+y^{2}+2 x+4 y+15=0\) would transform to
1 \(3 x^{2}+2 y^{2}-4 x+\overline{6 y+23}=0\)
2 \(3 x^{2}+y^{2}-4 x+6 y+21=0\)
3 \(3 x^{2}+y^{2}+4 x-6 y-21=0\)
4 \(3 x^{2}+y^{2}+4 x+6 y+21=0\)
Explanation:
(B): Given, The axis transformed to the point \((-1,1)\) Initial equation \(3 x^{2}+y^{2}+2 x+4 y+15=0\) We know that, The co-ordinate of transformed point \(\mathrm{x}=\mathrm{x}+\mathrm{h}\) \(\mathrm{y}=\mathrm{y}+\mathrm{k}\) \(\therefore \quad \mathrm{x}=\mathrm{x}-1\) Now, The equation transform to \(3(\mathrm{x}-1)^{2}+(\mathrm{y}+1)^{2}+2(\mathrm{x}-1)+4(\mathrm{y}+1)+15=0\) \(3\left(\mathrm{x}^{2}+1-2 \mathrm{x}\right)+\mathrm{y}^{2}+1+2 \mathrm{y}+2 \mathrm{x}-2+4 \mathrm{y}+4+15=0\) \(3 \mathrm{x}^{2}+3-6 \mathrm{x}+\mathrm{y}^{2}+1+2 \mathrm{y}+2 \mathrm{x}-2+4 \mathrm{y}+4+15=0\) \(3 x^{2}-4 x+y^{2}+6 y+21=0\) \(3 x^{2}+y^{2}-4 x+6 y+21=0\)
APEAPCET-2021-20.08.2021
Co-Ordinate system
88213
When the axes are rotated through an angle \(45^{\circ}\). the new coordinates of a point \(P\) are \((1,-1)\). The coordinates of \(P\) in the original system are
1 \((\sqrt{2}, \sqrt{2)}\)
2 \((\sqrt{2}, 0)\)
3 \(0, \sqrt{2})\)
4 \((-\sqrt{2}, 0)\)
Explanation:
(B) : Given, The angle of rotation of axis \(=45^{\circ}\) The new co-ordinates of a point \(\mathrm{P}=(1,-1)\) We know that, \(x=x \cos \theta-y \sin \theta\) \(y=x \sin \theta+y \cos \theta\) Now, \(x=1 \times \cos 45^{\circ}-(-1) \sin 45^{\circ}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\) \(y=1 \times \sin 45^{\circ}+(-1) \cos 45^{\circ}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\) \(\therefore \quad\) Original co-ordinate \(=(\sqrt{2}, 0)\)
88210
If the axis are rotated through an angle of \(30^{\circ}\) in the clockwise direction, the point \((4,2 \sqrt{3})\) in the new system is
1 \((2,3)\)
2 \((2, \sqrt{3})\)
3 \((\sqrt{3}, 2)\)
4 \((\sqrt{3}, 5)\)
Explanation:
(D) : Given, The coordinate of \((x, y)\) is \((4,2 \sqrt{3})\) \(x=4, y=2 \sqrt{3}\) \(\theta=-30^{\circ} \text { [Clock wise] }\) We know, \(\mathrm{X}=\mathrm{x} \cos \theta+\mathrm{y} \sin \theta\) and, \(\mathrm{Y}=\mathrm{x} \sin \theta+\mathrm{y} \cos \theta\) \(\therefore \quad \mathrm{X}=4 \cos 30^{\circ}-2 \sqrt{3} \sin 30^{\circ}\) \(=4 \times \frac{\sqrt{3}}{2}-2 \sqrt{3} \times \frac{1}{2}=\sqrt{3}\) \(Y=4 \sin 30^{\circ}+2 \sqrt{3} \cos 30^{\circ}\) \(=4 \times \frac{1}{2}+2 \sqrt{3} \times \frac{\sqrt{3}}{2}=2+3 \quad=5\) The given Point in the new system \((\sqrt{3}, 5)\)
JCECE-2015
Co-Ordinate system
88211
If the axis are rotated through an angle \(45^{\circ}\) in the positive direction without changing the origin, then the co-ordinates of the point \((\sqrt{2}, 4)\) in the old system are
1 \((1-2 \sqrt{2}, 1+2 \sqrt{2})\)
2 \((1+2 \sqrt{2}, 1-2 \sqrt{2})\)
3 \((2 \sqrt{2}, \sqrt{2})\)
4 \((\sqrt{2}, 2)\)
Explanation:
(A) : Given, The axis are rotated through an angle \(=45^{\circ}\) The co-ordinate of the point \(=(\sqrt{2}, 4)\) We know that, \(\mathrm{x}^{\prime}=\mathrm{x} \cos \theta-\mathrm{y} \sin \theta\) \(y^{\prime}=x \sin \theta+y \cos \theta\) Now, \(x^{\prime}=x \cos 45^{\circ}-y \sin 45^{\circ}\) \(x^{\prime}=\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}}-\frac{4}{\sqrt{2}}=1-2 \sqrt{2}\) \(y^{\prime}=x \sin \theta+y \cos \theta\) \(y^{\prime}=\sqrt{2} \sin 45^{\circ}+4 \cos 45^{\circ}\) \(y^{\prime}=\frac{\sqrt{2}}{\sqrt{2}}+\frac{4}{\sqrt{2}}\) \(y^{\prime}=1+2 \sqrt{2}\) \(\therefore\) The co-ordinate is \(\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)=(1-2 \sqrt{2}, 1+2 \sqrt{2})\)
AP EAMCET-2002
Co-Ordinate system
88212
If the axis are transformed to the point \((-1,1)\) then the equation \(3 x^{2}+y^{2}+2 x+4 y+15=0\) would transform to
1 \(3 x^{2}+2 y^{2}-4 x+\overline{6 y+23}=0\)
2 \(3 x^{2}+y^{2}-4 x+6 y+21=0\)
3 \(3 x^{2}+y^{2}+4 x-6 y-21=0\)
4 \(3 x^{2}+y^{2}+4 x+6 y+21=0\)
Explanation:
(B): Given, The axis transformed to the point \((-1,1)\) Initial equation \(3 x^{2}+y^{2}+2 x+4 y+15=0\) We know that, The co-ordinate of transformed point \(\mathrm{x}=\mathrm{x}+\mathrm{h}\) \(\mathrm{y}=\mathrm{y}+\mathrm{k}\) \(\therefore \quad \mathrm{x}=\mathrm{x}-1\) Now, The equation transform to \(3(\mathrm{x}-1)^{2}+(\mathrm{y}+1)^{2}+2(\mathrm{x}-1)+4(\mathrm{y}+1)+15=0\) \(3\left(\mathrm{x}^{2}+1-2 \mathrm{x}\right)+\mathrm{y}^{2}+1+2 \mathrm{y}+2 \mathrm{x}-2+4 \mathrm{y}+4+15=0\) \(3 \mathrm{x}^{2}+3-6 \mathrm{x}+\mathrm{y}^{2}+1+2 \mathrm{y}+2 \mathrm{x}-2+4 \mathrm{y}+4+15=0\) \(3 x^{2}-4 x+y^{2}+6 y+21=0\) \(3 x^{2}+y^{2}-4 x+6 y+21=0\)
APEAPCET-2021-20.08.2021
Co-Ordinate system
88213
When the axes are rotated through an angle \(45^{\circ}\). the new coordinates of a point \(P\) are \((1,-1)\). The coordinates of \(P\) in the original system are
1 \((\sqrt{2}, \sqrt{2)}\)
2 \((\sqrt{2}, 0)\)
3 \(0, \sqrt{2})\)
4 \((-\sqrt{2}, 0)\)
Explanation:
(B) : Given, The angle of rotation of axis \(=45^{\circ}\) The new co-ordinates of a point \(\mathrm{P}=(1,-1)\) We know that, \(x=x \cos \theta-y \sin \theta\) \(y=x \sin \theta+y \cos \theta\) Now, \(x=1 \times \cos 45^{\circ}-(-1) \sin 45^{\circ}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\) \(y=1 \times \sin 45^{\circ}+(-1) \cos 45^{\circ}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\) \(\therefore \quad\) Original co-ordinate \(=(\sqrt{2}, 0)\)
88210
If the axis are rotated through an angle of \(30^{\circ}\) in the clockwise direction, the point \((4,2 \sqrt{3})\) in the new system is
1 \((2,3)\)
2 \((2, \sqrt{3})\)
3 \((\sqrt{3}, 2)\)
4 \((\sqrt{3}, 5)\)
Explanation:
(D) : Given, The coordinate of \((x, y)\) is \((4,2 \sqrt{3})\) \(x=4, y=2 \sqrt{3}\) \(\theta=-30^{\circ} \text { [Clock wise] }\) We know, \(\mathrm{X}=\mathrm{x} \cos \theta+\mathrm{y} \sin \theta\) and, \(\mathrm{Y}=\mathrm{x} \sin \theta+\mathrm{y} \cos \theta\) \(\therefore \quad \mathrm{X}=4 \cos 30^{\circ}-2 \sqrt{3} \sin 30^{\circ}\) \(=4 \times \frac{\sqrt{3}}{2}-2 \sqrt{3} \times \frac{1}{2}=\sqrt{3}\) \(Y=4 \sin 30^{\circ}+2 \sqrt{3} \cos 30^{\circ}\) \(=4 \times \frac{1}{2}+2 \sqrt{3} \times \frac{\sqrt{3}}{2}=2+3 \quad=5\) The given Point in the new system \((\sqrt{3}, 5)\)
JCECE-2015
Co-Ordinate system
88211
If the axis are rotated through an angle \(45^{\circ}\) in the positive direction without changing the origin, then the co-ordinates of the point \((\sqrt{2}, 4)\) in the old system are
1 \((1-2 \sqrt{2}, 1+2 \sqrt{2})\)
2 \((1+2 \sqrt{2}, 1-2 \sqrt{2})\)
3 \((2 \sqrt{2}, \sqrt{2})\)
4 \((\sqrt{2}, 2)\)
Explanation:
(A) : Given, The axis are rotated through an angle \(=45^{\circ}\) The co-ordinate of the point \(=(\sqrt{2}, 4)\) We know that, \(\mathrm{x}^{\prime}=\mathrm{x} \cos \theta-\mathrm{y} \sin \theta\) \(y^{\prime}=x \sin \theta+y \cos \theta\) Now, \(x^{\prime}=x \cos 45^{\circ}-y \sin 45^{\circ}\) \(x^{\prime}=\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}}-\frac{4}{\sqrt{2}}=1-2 \sqrt{2}\) \(y^{\prime}=x \sin \theta+y \cos \theta\) \(y^{\prime}=\sqrt{2} \sin 45^{\circ}+4 \cos 45^{\circ}\) \(y^{\prime}=\frac{\sqrt{2}}{\sqrt{2}}+\frac{4}{\sqrt{2}}\) \(y^{\prime}=1+2 \sqrt{2}\) \(\therefore\) The co-ordinate is \(\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)=(1-2 \sqrt{2}, 1+2 \sqrt{2})\)
AP EAMCET-2002
Co-Ordinate system
88212
If the axis are transformed to the point \((-1,1)\) then the equation \(3 x^{2}+y^{2}+2 x+4 y+15=0\) would transform to
1 \(3 x^{2}+2 y^{2}-4 x+\overline{6 y+23}=0\)
2 \(3 x^{2}+y^{2}-4 x+6 y+21=0\)
3 \(3 x^{2}+y^{2}+4 x-6 y-21=0\)
4 \(3 x^{2}+y^{2}+4 x+6 y+21=0\)
Explanation:
(B): Given, The axis transformed to the point \((-1,1)\) Initial equation \(3 x^{2}+y^{2}+2 x+4 y+15=0\) We know that, The co-ordinate of transformed point \(\mathrm{x}=\mathrm{x}+\mathrm{h}\) \(\mathrm{y}=\mathrm{y}+\mathrm{k}\) \(\therefore \quad \mathrm{x}=\mathrm{x}-1\) Now, The equation transform to \(3(\mathrm{x}-1)^{2}+(\mathrm{y}+1)^{2}+2(\mathrm{x}-1)+4(\mathrm{y}+1)+15=0\) \(3\left(\mathrm{x}^{2}+1-2 \mathrm{x}\right)+\mathrm{y}^{2}+1+2 \mathrm{y}+2 \mathrm{x}-2+4 \mathrm{y}+4+15=0\) \(3 \mathrm{x}^{2}+3-6 \mathrm{x}+\mathrm{y}^{2}+1+2 \mathrm{y}+2 \mathrm{x}-2+4 \mathrm{y}+4+15=0\) \(3 x^{2}-4 x+y^{2}+6 y+21=0\) \(3 x^{2}+y^{2}-4 x+6 y+21=0\)
APEAPCET-2021-20.08.2021
Co-Ordinate system
88213
When the axes are rotated through an angle \(45^{\circ}\). the new coordinates of a point \(P\) are \((1,-1)\). The coordinates of \(P\) in the original system are
1 \((\sqrt{2}, \sqrt{2)}\)
2 \((\sqrt{2}, 0)\)
3 \(0, \sqrt{2})\)
4 \((-\sqrt{2}, 0)\)
Explanation:
(B) : Given, The angle of rotation of axis \(=45^{\circ}\) The new co-ordinates of a point \(\mathrm{P}=(1,-1)\) We know that, \(x=x \cos \theta-y \sin \theta\) \(y=x \sin \theta+y \cos \theta\) Now, \(x=1 \times \cos 45^{\circ}-(-1) \sin 45^{\circ}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\) \(y=1 \times \sin 45^{\circ}+(-1) \cos 45^{\circ}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\) \(\therefore \quad\) Original co-ordinate \(=(\sqrt{2}, 0)\)
