Question

Point Q with coordinates (4, –7) is rotated 270° clockwise about (5, 1) . What are the coordinates of its image?

Answer 1

Answer:

The coordinates of the image are $Q'(x,y) = (13,0)$.

Step-by-step explanation:

The rotation of point Q around a given point of reference is defined by:

$Q'(x,y) = O(x,y) + r_{OQ}\cdot (\cos (\theta_{OQ}\pm \theta_{R}),\sin (\theta_{OQ}\pm \theta_{R}))$ (1)

Where:

$O(x,y)$ - Point of reference.

$r_{OQ}$ - Magnitude of line segment OP.

$\theta_{OQ}$ - Direction of the point Q with respect to the point of reference, measured in sexagesimal degrees.

$\theta_{R}$ - Rotation angle, measured in sexagesimal degrees.

$Q'(x,y)$ - Rotated point.

Please notice that positive sign means counterclockwise rotation, whereas the negative sign represents clockwise rotation.

The magnitude and direction of the line segment are defined by the following expressions:

Magnitude

$r_{OQ} = \sqrt{(x_{Q}-x_{O})^{2}+(y_{Q}-y_{O})^{2}}$ (2)

Direction

$\theta_{OQ} = \tan^{-1}\left(\frac{y_{Q}-y_{O}}{x_{Q}-x_{O}} \right)$ (4)

Where:

$x_{O}$, $y_{O}$ - Coordinates of point of reference.

$x_{Q}$, $y_{Q}$ - Coordinates of the point Q.

If we know that $x_{O} = 5$, $y_{O} = 1$, $x_{Q} = 4$, $y_{Q} = -7$, then the magnitude and direction of the segment OQ are, respectively:

$r_{OQ} = \sqrt{(4-5)^{2}+(-7-1)^{2}}$

$r_{OQ} \approx 8.062$

$\theta_{OQ} = \tan^{-1}\left(\frac{-7-1}{4-5} \right)$

$\theta_{OQ} = \tan^{-1}\left(\frac{-8}{-1} \right)$ (Which means that line segment OQ belongs to third quadrant)

$\theta_{OQ} \approx 262.875^{\circ}$

If we know that rotation is clockwise,  $x_{O} = 5$, $y_{O} = 1$, $r_{OQ} \approx 8.062$ and $\theta_{OQ} \approx 262.875^{\circ}$, then the rotated point is:

$Q'(x,y) = (5,1) + (8.062\cdot \cos (262.875^{\circ}-270^{\circ}),8.062\cdot \sin (262.875^{\circ}-270^{\circ}))$

$Q'(x,y) = (5,1) + (8, -1)$

$Q'(x,y) = (13, 0)$

The coordinates of the image are $Q'(x,y) = (13,0)$.