NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88219
The point \((2,3)\) is first reflected in the straight line \(y=x\) and then translated through a distance of 2 units along the positive direction \(\mathrm{X}\)-axis. The coordinates of the transformed point are
1 \((5,4)\)
2 \((2,3)\)
3 \((5,2)\)
4 \((4,5)\)
Explanation:
(C) : Given, The coordinate of the point is \((2,3)\) We know that, The coordinate reflect in the straight line \(y=x\) \(\therefore\) The coordinate of reflected point is \(\mathrm{P}_{2}(3,2)\) Now, The point \(\mathrm{P}_{2}\) translated in the positive direction \(\mathrm{x}-\) axis by 2 units \(\therefore\) The New coordinate after translation is \(P_{3}=(3+2,2) \Rightarrow P_{3}=(5,2)\)
AP EAMCET-2015
Co-Ordinate system
88220
If the axis are rotated through an angle \(45^{\circ}\), the coordinates of the point \((2 \sqrt{2},-3 \sqrt{2})\) in the new system are
88221
Without changing the direction of the axis, the origin is transferred to the point \((2,3)\). Then the equation \(x^{2}+y^{2}-4 x-6 y+9=0\) changes to
(B) : Given, The equation is \(x^{2}+y^{2}-4 x-6 y+9=0\) The origin is transferred to the point \((2,3)\) We know that, \(\mathrm{x}^{\prime}=\mathrm{x}+2\) \(\mathrm{y}^{\prime}=\mathrm{y}+3\) Now, \((x+2)^{2}+(y+3)^{2}-4(x+2)-6(y+3)+9=0\) \(x^{2}+4+4 x+y^{2}+9+6 y-4 x-8-6 y-18+9=0\) \(x^{2}+y^{2}-4=0\)
WB JEE-2018
Co-Ordinate system
88222
If the coordinates of a point \(P\) changes to \((2,-6)\) when the coordinate axis are rotated through an angle of \(135^{\circ}\), then the coordinates of \(P\) in the original system are
1 \((-2,6)\)
2 \((-6,2)\)
3 \((2 \sqrt{2,} 4 \sqrt{2})\)
4 \((\sqrt{2,}-\sqrt{2})\)
Explanation:
(C) : Given, The coordinated of change point \(\mathrm{P}(2,-6)\) Angle of rotation \((\theta)=135^{\circ}\) We know that, \(x=x \cos \theta-y \sin \theta\) So, \(\quad \begin{array}{ll}y=y \cos \theta+x \sin \theta \\ x =2 \cos 135^{\circ}-(-6) \sin 135\end{array}\) \(x=2\left(-\frac{1}{\sqrt{2}}\right)+6\left(\frac{1}{\sqrt{2}}\right)=\frac{4}{\sqrt{2}}=2 \sqrt{2}\) And, \(\mathrm{y}=\mathrm{Y} \cos \theta+\mathrm{X} \sin \theta\) \(y=(-6)\left(-\frac{1}{\sqrt{2}}\right)-2\left(\frac{1}{\sqrt{2}}\right) \Rightarrow y=\frac{6}{\sqrt{2}}+\frac{2}{\sqrt{2}}\) \(=\frac{8}{\sqrt{2}}=4 \sqrt{2}\) \(\therefore\) The original coordinate \((2 \sqrt{2}, 4 \sqrt{2})\)
88219
The point \((2,3)\) is first reflected in the straight line \(y=x\) and then translated through a distance of 2 units along the positive direction \(\mathrm{X}\)-axis. The coordinates of the transformed point are
1 \((5,4)\)
2 \((2,3)\)
3 \((5,2)\)
4 \((4,5)\)
Explanation:
(C) : Given, The coordinate of the point is \((2,3)\) We know that, The coordinate reflect in the straight line \(y=x\) \(\therefore\) The coordinate of reflected point is \(\mathrm{P}_{2}(3,2)\) Now, The point \(\mathrm{P}_{2}\) translated in the positive direction \(\mathrm{x}-\) axis by 2 units \(\therefore\) The New coordinate after translation is \(P_{3}=(3+2,2) \Rightarrow P_{3}=(5,2)\)
AP EAMCET-2015
Co-Ordinate system
88220
If the axis are rotated through an angle \(45^{\circ}\), the coordinates of the point \((2 \sqrt{2},-3 \sqrt{2})\) in the new system are
88221
Without changing the direction of the axis, the origin is transferred to the point \((2,3)\). Then the equation \(x^{2}+y^{2}-4 x-6 y+9=0\) changes to
(B) : Given, The equation is \(x^{2}+y^{2}-4 x-6 y+9=0\) The origin is transferred to the point \((2,3)\) We know that, \(\mathrm{x}^{\prime}=\mathrm{x}+2\) \(\mathrm{y}^{\prime}=\mathrm{y}+3\) Now, \((x+2)^{2}+(y+3)^{2}-4(x+2)-6(y+3)+9=0\) \(x^{2}+4+4 x+y^{2}+9+6 y-4 x-8-6 y-18+9=0\) \(x^{2}+y^{2}-4=0\)
WB JEE-2018
Co-Ordinate system
88222
If the coordinates of a point \(P\) changes to \((2,-6)\) when the coordinate axis are rotated through an angle of \(135^{\circ}\), then the coordinates of \(P\) in the original system are
1 \((-2,6)\)
2 \((-6,2)\)
3 \((2 \sqrt{2,} 4 \sqrt{2})\)
4 \((\sqrt{2,}-\sqrt{2})\)
Explanation:
(C) : Given, The coordinated of change point \(\mathrm{P}(2,-6)\) Angle of rotation \((\theta)=135^{\circ}\) We know that, \(x=x \cos \theta-y \sin \theta\) So, \(\quad \begin{array}{ll}y=y \cos \theta+x \sin \theta \\ x =2 \cos 135^{\circ}-(-6) \sin 135\end{array}\) \(x=2\left(-\frac{1}{\sqrt{2}}\right)+6\left(\frac{1}{\sqrt{2}}\right)=\frac{4}{\sqrt{2}}=2 \sqrt{2}\) And, \(\mathrm{y}=\mathrm{Y} \cos \theta+\mathrm{X} \sin \theta\) \(y=(-6)\left(-\frac{1}{\sqrt{2}}\right)-2\left(\frac{1}{\sqrt{2}}\right) \Rightarrow y=\frac{6}{\sqrt{2}}+\frac{2}{\sqrt{2}}\) \(=\frac{8}{\sqrt{2}}=4 \sqrt{2}\) \(\therefore\) The original coordinate \((2 \sqrt{2}, 4 \sqrt{2})\)
88219
The point \((2,3)\) is first reflected in the straight line \(y=x\) and then translated through a distance of 2 units along the positive direction \(\mathrm{X}\)-axis. The coordinates of the transformed point are
1 \((5,4)\)
2 \((2,3)\)
3 \((5,2)\)
4 \((4,5)\)
Explanation:
(C) : Given, The coordinate of the point is \((2,3)\) We know that, The coordinate reflect in the straight line \(y=x\) \(\therefore\) The coordinate of reflected point is \(\mathrm{P}_{2}(3,2)\) Now, The point \(\mathrm{P}_{2}\) translated in the positive direction \(\mathrm{x}-\) axis by 2 units \(\therefore\) The New coordinate after translation is \(P_{3}=(3+2,2) \Rightarrow P_{3}=(5,2)\)
AP EAMCET-2015
Co-Ordinate system
88220
If the axis are rotated through an angle \(45^{\circ}\), the coordinates of the point \((2 \sqrt{2},-3 \sqrt{2})\) in the new system are
88221
Without changing the direction of the axis, the origin is transferred to the point \((2,3)\). Then the equation \(x^{2}+y^{2}-4 x-6 y+9=0\) changes to
(B) : Given, The equation is \(x^{2}+y^{2}-4 x-6 y+9=0\) The origin is transferred to the point \((2,3)\) We know that, \(\mathrm{x}^{\prime}=\mathrm{x}+2\) \(\mathrm{y}^{\prime}=\mathrm{y}+3\) Now, \((x+2)^{2}+(y+3)^{2}-4(x+2)-6(y+3)+9=0\) \(x^{2}+4+4 x+y^{2}+9+6 y-4 x-8-6 y-18+9=0\) \(x^{2}+y^{2}-4=0\)
WB JEE-2018
Co-Ordinate system
88222
If the coordinates of a point \(P\) changes to \((2,-6)\) when the coordinate axis are rotated through an angle of \(135^{\circ}\), then the coordinates of \(P\) in the original system are
1 \((-2,6)\)
2 \((-6,2)\)
3 \((2 \sqrt{2,} 4 \sqrt{2})\)
4 \((\sqrt{2,}-\sqrt{2})\)
Explanation:
(C) : Given, The coordinated of change point \(\mathrm{P}(2,-6)\) Angle of rotation \((\theta)=135^{\circ}\) We know that, \(x=x \cos \theta-y \sin \theta\) So, \(\quad \begin{array}{ll}y=y \cos \theta+x \sin \theta \\ x =2 \cos 135^{\circ}-(-6) \sin 135\end{array}\) \(x=2\left(-\frac{1}{\sqrt{2}}\right)+6\left(\frac{1}{\sqrt{2}}\right)=\frac{4}{\sqrt{2}}=2 \sqrt{2}\) And, \(\mathrm{y}=\mathrm{Y} \cos \theta+\mathrm{X} \sin \theta\) \(y=(-6)\left(-\frac{1}{\sqrt{2}}\right)-2\left(\frac{1}{\sqrt{2}}\right) \Rightarrow y=\frac{6}{\sqrt{2}}+\frac{2}{\sqrt{2}}\) \(=\frac{8}{\sqrt{2}}=4 \sqrt{2}\) \(\therefore\) The original coordinate \((2 \sqrt{2}, 4 \sqrt{2})\)
88219
The point \((2,3)\) is first reflected in the straight line \(y=x\) and then translated through a distance of 2 units along the positive direction \(\mathrm{X}\)-axis. The coordinates of the transformed point are
1 \((5,4)\)
2 \((2,3)\)
3 \((5,2)\)
4 \((4,5)\)
Explanation:
(C) : Given, The coordinate of the point is \((2,3)\) We know that, The coordinate reflect in the straight line \(y=x\) \(\therefore\) The coordinate of reflected point is \(\mathrm{P}_{2}(3,2)\) Now, The point \(\mathrm{P}_{2}\) translated in the positive direction \(\mathrm{x}-\) axis by 2 units \(\therefore\) The New coordinate after translation is \(P_{3}=(3+2,2) \Rightarrow P_{3}=(5,2)\)
AP EAMCET-2015
Co-Ordinate system
88220
If the axis are rotated through an angle \(45^{\circ}\), the coordinates of the point \((2 \sqrt{2},-3 \sqrt{2})\) in the new system are
88221
Without changing the direction of the axis, the origin is transferred to the point \((2,3)\). Then the equation \(x^{2}+y^{2}-4 x-6 y+9=0\) changes to
(B) : Given, The equation is \(x^{2}+y^{2}-4 x-6 y+9=0\) The origin is transferred to the point \((2,3)\) We know that, \(\mathrm{x}^{\prime}=\mathrm{x}+2\) \(\mathrm{y}^{\prime}=\mathrm{y}+3\) Now, \((x+2)^{2}+(y+3)^{2}-4(x+2)-6(y+3)+9=0\) \(x^{2}+4+4 x+y^{2}+9+6 y-4 x-8-6 y-18+9=0\) \(x^{2}+y^{2}-4=0\)
WB JEE-2018
Co-Ordinate system
88222
If the coordinates of a point \(P\) changes to \((2,-6)\) when the coordinate axis are rotated through an angle of \(135^{\circ}\), then the coordinates of \(P\) in the original system are
1 \((-2,6)\)
2 \((-6,2)\)
3 \((2 \sqrt{2,} 4 \sqrt{2})\)
4 \((\sqrt{2,}-\sqrt{2})\)
Explanation:
(C) : Given, The coordinated of change point \(\mathrm{P}(2,-6)\) Angle of rotation \((\theta)=135^{\circ}\) We know that, \(x=x \cos \theta-y \sin \theta\) So, \(\quad \begin{array}{ll}y=y \cos \theta+x \sin \theta \\ x =2 \cos 135^{\circ}-(-6) \sin 135\end{array}\) \(x=2\left(-\frac{1}{\sqrt{2}}\right)+6\left(\frac{1}{\sqrt{2}}\right)=\frac{4}{\sqrt{2}}=2 \sqrt{2}\) And, \(\mathrm{y}=\mathrm{Y} \cos \theta+\mathrm{X} \sin \theta\) \(y=(-6)\left(-\frac{1}{\sqrt{2}}\right)-2\left(\frac{1}{\sqrt{2}}\right) \Rightarrow y=\frac{6}{\sqrt{2}}+\frac{2}{\sqrt{2}}\) \(=\frac{8}{\sqrt{2}}=4 \sqrt{2}\) \(\therefore\) The original coordinate \((2 \sqrt{2}, 4 \sqrt{2})\)
