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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 $X$ -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 $y = x$

$∴$ The coordinate of reflected point is $P_{2} (3, 2)$
Now, The point $P_{2}$ translated in the positive direction $x -$ axis by 2 units
$∴$ 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

1

(3 \sqrt{3}, - 5)

2

(- 1, - 5)

3

(5 \sqrt{3}, - 7)

4

(7, - \sqrt{3})

Explanation:

(B): Given,
Coordinate $(x, y) = (2 \sqrt{2}, - 3 \sqrt{2})$
Let new coordinates be ( $x^{'}, y^{'})$ . Then,
$x^{'} = x \cos θ + y \cos θ$
And $y^{'} = - x xin θ + y \cos θ$
Here, $θ = 45^{\circ}$
$∴ x^{'} = 2 \sqrt{2} \cos 45^{\circ} + (- 3 \sqrt{2}) \sin 45^{\circ}$
$= 2 \sqrt{2} \times \frac{1}{\sqrt{2}} - 3 \sqrt{2} \times \frac{1}{\sqrt{2}}$
$= 2 - 3 = - 1$
And, $y^{'} = - 2 \sqrt{2} \sin 45^{\circ} + (- 3 \sqrt{2}) \cos 45^{\circ}$
$= - 2 \sqrt{2} \times \frac{1}{\sqrt{2}} - 3 \sqrt{2} \times \frac{1}{\sqrt{2}} = - 2 - 3 = - 5$
New, coordinate is $(- 1, - 5)$ .

AP EAMCET-2021-19.08.2021

Co-Ordinate system

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

1

x^{2} + y^{2} + 4 = 0

2

x^{2} + y^{2} = 4

3

x^{2} + y^{2} - 8 x - 12 y + 48 = 0

4

x^{2} + y^{2} = 9

Explanation:

(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,
$x^{'} = x + 2$
$y^{'} = y + 3$
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 $P (2, - 6)$
Angle of rotation $(θ) = 135^{\circ}$
We know that,
$x = x \cos θ - y \sin θ$
So, $\begin{array}{ll} y = y \cos θ + x \sin θ \\ x = 2 \cos 135^{\circ} - (- 6) \sin 135 \end{array}$
$x = 2 (- \frac{1}{\sqrt{2}}) + 6 (\frac{1}{\sqrt{2}}) = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$
And, $y = Y \cos θ + X \sin θ$
$y = (- 6) (- \frac{1}{\sqrt{2}}) - 2 (\frac{1}{\sqrt{2}}) \Rightarrow y = \frac{6}{\sqrt{2}} + \frac{2}{\sqrt{2}}$
$= \frac{8}{\sqrt{2}} = 4 \sqrt{2}$
$∴$ The original coordinate $(2 \sqrt{2}, 4 \sqrt{2})$

AP EAMCET-2018-22.04.2018

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 $X$ -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 $y = x$

$∴$ The coordinate of reflected point is $P_{2} (3, 2)$
Now, The point $P_{2}$ translated in the positive direction $x -$ axis by 2 units
$∴$ 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

1

(3 \sqrt{3}, - 5)

2

(- 1, - 5)

3

(5 \sqrt{3}, - 7)

4

(7, - \sqrt{3})

Explanation:

(B): Given,
Coordinate $(x, y) = (2 \sqrt{2}, - 3 \sqrt{2})$
Let new coordinates be ( $x^{'}, y^{'})$ . Then,
$x^{'} = x \cos θ + y \cos θ$
And $y^{'} = - x xin θ + y \cos θ$
Here, $θ = 45^{\circ}$
$∴ x^{'} = 2 \sqrt{2} \cos 45^{\circ} + (- 3 \sqrt{2}) \sin 45^{\circ}$
$= 2 \sqrt{2} \times \frac{1}{\sqrt{2}} - 3 \sqrt{2} \times \frac{1}{\sqrt{2}}$
$= 2 - 3 = - 1$
And, $y^{'} = - 2 \sqrt{2} \sin 45^{\circ} + (- 3 \sqrt{2}) \cos 45^{\circ}$
$= - 2 \sqrt{2} \times \frac{1}{\sqrt{2}} - 3 \sqrt{2} \times \frac{1}{\sqrt{2}} = - 2 - 3 = - 5$
New, coordinate is $(- 1, - 5)$ .

AP EAMCET-2021-19.08.2021

Co-Ordinate system

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

1

x^{2} + y^{2} + 4 = 0

2

x^{2} + y^{2} = 4

3

x^{2} + y^{2} - 8 x - 12 y + 48 = 0

4

x^{2} + y^{2} = 9

Explanation:

(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,
$x^{'} = x + 2$
$y^{'} = y + 3$
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 $P (2, - 6)$
Angle of rotation $(θ) = 135^{\circ}$
We know that,
$x = x \cos θ - y \sin θ$
So, $\begin{array}{ll} y = y \cos θ + x \sin θ \\ x = 2 \cos 135^{\circ} - (- 6) \sin 135 \end{array}$
$x = 2 (- \frac{1}{\sqrt{2}}) + 6 (\frac{1}{\sqrt{2}}) = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$
And, $y = Y \cos θ + X \sin θ$
$y = (- 6) (- \frac{1}{\sqrt{2}}) - 2 (\frac{1}{\sqrt{2}}) \Rightarrow y = \frac{6}{\sqrt{2}} + \frac{2}{\sqrt{2}}$
$= \frac{8}{\sqrt{2}} = 4 \sqrt{2}$
$∴$ The original coordinate $(2 \sqrt{2}, 4 \sqrt{2})$

AP EAMCET-2018-22.04.2018

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 $X$ -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 $y = x$

$∴$ The coordinate of reflected point is $P_{2} (3, 2)$
Now, The point $P_{2}$ translated in the positive direction $x -$ axis by 2 units
$∴$ 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

1

(3 \sqrt{3}, - 5)

2

(- 1, - 5)

3

(5 \sqrt{3}, - 7)

4

(7, - \sqrt{3})

Explanation:

(B): Given,
Coordinate $(x, y) = (2 \sqrt{2}, - 3 \sqrt{2})$
Let new coordinates be ( $x^{'}, y^{'})$ . Then,
$x^{'} = x \cos θ + y \cos θ$
And $y^{'} = - x xin θ + y \cos θ$
Here, $θ = 45^{\circ}$
$∴ x^{'} = 2 \sqrt{2} \cos 45^{\circ} + (- 3 \sqrt{2}) \sin 45^{\circ}$
$= 2 \sqrt{2} \times \frac{1}{\sqrt{2}} - 3 \sqrt{2} \times \frac{1}{\sqrt{2}}$
$= 2 - 3 = - 1$
And, $y^{'} = - 2 \sqrt{2} \sin 45^{\circ} + (- 3 \sqrt{2}) \cos 45^{\circ}$
$= - 2 \sqrt{2} \times \frac{1}{\sqrt{2}} - 3 \sqrt{2} \times \frac{1}{\sqrt{2}} = - 2 - 3 = - 5$
New, coordinate is $(- 1, - 5)$ .

AP EAMCET-2021-19.08.2021

Co-Ordinate system

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

1

x^{2} + y^{2} + 4 = 0

2

x^{2} + y^{2} = 4

3

x^{2} + y^{2} - 8 x - 12 y + 48 = 0

4

x^{2} + y^{2} = 9

Explanation:

(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,
$x^{'} = x + 2$
$y^{'} = y + 3$
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 $P (2, - 6)$
Angle of rotation $(θ) = 135^{\circ}$
We know that,
$x = x \cos θ - y \sin θ$
So, $\begin{array}{ll} y = y \cos θ + x \sin θ \\ x = 2 \cos 135^{\circ} - (- 6) \sin 135 \end{array}$
$x = 2 (- \frac{1}{\sqrt{2}}) + 6 (\frac{1}{\sqrt{2}}) = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$
And, $y = Y \cos θ + X \sin θ$
$y = (- 6) (- \frac{1}{\sqrt{2}}) - 2 (\frac{1}{\sqrt{2}}) \Rightarrow y = \frac{6}{\sqrt{2}} + \frac{2}{\sqrt{2}}$
$= \frac{8}{\sqrt{2}} = 4 \sqrt{2}$
$∴$ The original coordinate $(2 \sqrt{2}, 4 \sqrt{2})$

AP EAMCET-2018-22.04.2018

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 $X$ -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 $y = x$

$∴$ The coordinate of reflected point is $P_{2} (3, 2)$
Now, The point $P_{2}$ translated in the positive direction $x -$ axis by 2 units
$∴$ 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

1

(3 \sqrt{3}, - 5)

2

(- 1, - 5)

3

(5 \sqrt{3}, - 7)

4

(7, - \sqrt{3})

Explanation:

(B): Given,
Coordinate $(x, y) = (2 \sqrt{2}, - 3 \sqrt{2})$
Let new coordinates be ( $x^{'}, y^{'})$ . Then,
$x^{'} = x \cos θ + y \cos θ$
And $y^{'} = - x xin θ + y \cos θ$
Here, $θ = 45^{\circ}$
$∴ x^{'} = 2 \sqrt{2} \cos 45^{\circ} + (- 3 \sqrt{2}) \sin 45^{\circ}$
$= 2 \sqrt{2} \times \frac{1}{\sqrt{2}} - 3 \sqrt{2} \times \frac{1}{\sqrt{2}}$
$= 2 - 3 = - 1$
And, $y^{'} = - 2 \sqrt{2} \sin 45^{\circ} + (- 3 \sqrt{2}) \cos 45^{\circ}$
$= - 2 \sqrt{2} \times \frac{1}{\sqrt{2}} - 3 \sqrt{2} \times \frac{1}{\sqrt{2}} = - 2 - 3 = - 5$
New, coordinate is $(- 1, - 5)$ .

AP EAMCET-2021-19.08.2021

Co-Ordinate system

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

1

x^{2} + y^{2} + 4 = 0

2

x^{2} + y^{2} = 4

3

x^{2} + y^{2} - 8 x - 12 y + 48 = 0

4

x^{2} + y^{2} = 9

Explanation:

(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,
$x^{'} = x + 2$
$y^{'} = y + 3$
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 $P (2, - 6)$
Angle of rotation $(θ) = 135^{\circ}$
We know that,
$x = x \cos θ - y \sin θ$
So, $\begin{array}{ll} y = y \cos θ + x \sin θ \\ x = 2 \cos 135^{\circ} - (- 6) \sin 135 \end{array}$
$x = 2 (- \frac{1}{\sqrt{2}}) + 6 (\frac{1}{\sqrt{2}}) = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$
And, $y = Y \cos θ + X \sin θ$
$y = (- 6) (- \frac{1}{\sqrt{2}}) - 2 (\frac{1}{\sqrt{2}}) \Rightarrow y = \frac{6}{\sqrt{2}} + \frac{2}{\sqrt{2}}$
$= \frac{8}{\sqrt{2}} = 4 \sqrt{2}$
$∴$ The original coordinate $(2 \sqrt{2}, 4 \sqrt{2})$

AP EAMCET-2018-22.04.2018

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