Example Question - coordinate transformation
Here are examples of questions we've helped users solve.
Dilation of a Point
The question is asking for the image of the point (12, −4) after a dilation by a scale factor of \( \frac{1}{3} \), centered at the origin. To find the image of a point after dilation, you multiply the coordinates of the original point by the scale factor. Here the scale factor is \( \frac{1}{3} \). Thus, the transformed coordinates (x', y') of the original point (x, y) = (12, −4) would be calculated by: x' = x * scale factor y' = y * scale factor Calculating these: x' = 12 * \( \frac{1}{3} \) = 4 y' = −4 * \( \frac{1}{3} \) = −\( \frac{4}{3} \) Therefore, the image of the point (12, −4) after the dilation is (4, −\( \frac{4}{3} \)).
Dilation of a Point
To perform a dilation on a point with a scale factor centered at the origin, you simply multiply the x and y coordinates of the point by the scale factor. For the point (6, 12) and a scale factor of \( \frac{1}{2} \), the calculation would be as follows: - The x-coordinate after dilation: \( 6 \times \frac{1}{2} = 3 \) - The y-coordinate after dilation: \( 12 \times \frac{1}{2} = 6 \) Therefore, the image of the point (6, 12) after a dilation with a scale factor of \( \frac{1}{2} \) centered at the origin is (3, 6).
Dilation of a Point
To find the image of the point (-8, -4) after a dilation by a scale factor of \( \frac{1}{4} \) centered at the origin, you multiply each coordinate of the original point by the scale factor. So, the new coordinates would be: \( x' = x \cdot \frac{1}{4} = -8 \cdot \frac{1}{4} = -2 \) \( y' = y \cdot \frac{1}{4} = -4 \cdot \frac{1}{4} = -1 \) Therefore, the image of the point (-8, -4) after the dilation is (-2, -1).
Dilation of a Point with Scale Factor and Origin
To find the image of a point after a dilation with a scale factor, you multiply each coordinate of the original point by the scale factor. When the dilation is centered at the origin (0, 0), this operation is straightforward. The original point is (-6, 7), and the scale factor is 5. Therefore, you multiply each coordinate of the point by the scale factor: New x-coordinate: -6 * 5 = -30 New y-coordinate: 7 * 5 = 35 So the image of the point (-6, 7) after dilation by a scale factor of 5, centered at the origin, is (-30, 35).
Reflecting a Point Over the Y-Axis
The image shows a question asking for the result of reflecting the point (-1, -8) over the y-axis. To reflect a point over the y-axis, you change the sign of the x-coordinate while the y-coordinate remains the same. Therefore, reflecting the point (-1, -8) over the y-axis will result in the point (1, -8). This new point has the original y-coordinate, with the x-coordinate being the opposite sign of the original. Hence, the image of (-1, -8) after a reflection over the y-axis is (1, -8).
Reflecting a Point over y = -x Line
To find the image of a point after a reflection over the line y = x, you would normally swap the x- and y-coordinates of the point. However, the line given in this question is y = -x, which means we need to swap the coordinates and change their signs. The point given is (-7, 5). After reflecting this point over the line y = -x, the x-coordinate becomes -5 and the y-coordinate becomes 7, because you swap them and change the signs. Hence, the image of point (-7, 5) after a reflection over the line y = -x is (-5, 7).
Rotating Points Around the Origin
To solve the question in the image, we need to determine the new positions of the vertices after a 270° counterclockwise rotation around the origin. The original coordinates of the vertices appear to be: - J(6,3) - K(9,3) - L(6,-2) Here are the general rules for rotating points around the origin: - For a 90° counterclockwise rotation, (x, y) becomes (-y, x). - For a 180° rotation, (x, y) becomes (-x, -y). - For a 270° counterclockwise rotation, (x, y) becomes (y, -x). Applying the rule for a 270° rotation to each point: For J(6,3): Rotate 270° counterclockwise to get J'(3,-6). For K(9,3): Rotate 270° counterclockwise to get K'(3,-9). For L(6,-2): Rotate 270° counterclockwise to get L'(-2,-6). Therefore, the new coordinates after the 270° counterclockwise rotation around the origin are: - J'(3, -6) - K'(3, -9) - L'(-2, -6)
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