ΔABC has coordinates of A (–6, –3), B (7, –7), and C (0, 3). Find the coordinates of its image after a dilation centered at the origin with a scale factor of 1.5.

A (–6, –3), B (7, –7), C (0, 3)

A (–9, –3), B (10.5, –7), C (0, 3)

A (–9, –4.5), B (10.5, –10.5), C (0, 4.5)

A (–4, –2), B (4.67, –4.67), C (0, 2)

1 answer

To find the coordinates of the image of triangle ΔABC after a dilation centered at the origin with a scale factor of 1.5, we can use the formula for dilation:

\[ (x', y') = (kx, ky) \]

where \((x, y)\) are the original coordinates, \(k\) is the scale factor, and \((x', y')\) are the new coordinates after dilation.

Given the scale factor \(k = 1.5\), we will apply this to each vertex of triangle ΔABC.

  1. For point A (–6, –3): \[ A' = (1.5 \cdot -6, 1.5 \cdot -3) = (-9, -4.5) \]

  2. For point B (7, –7): \[ B' = (1.5 \cdot 7, 1.5 \cdot -7) = (10.5, -10.5) \]

  3. For point C (0, 3): \[ C' = (1.5 \cdot 0, 1.5 \cdot 3) = (0, 4.5) \]

Thus, the coordinates of the image of triangle ΔABC after dilation are:

  • \(A' (-9, -4.5)\)
  • \(B' (10.5, -10.5)\)
  • \(C' (0, 4.5)\)

The correct answer is: A (–9, –4.5), B (10.5, –10.5), C (0, 4.5)