Triangle


is the image of

under a dilation with a scale factor of

.
A first quadrant coordinate plane. A triange has vertices A prime at zero, ten, B at four, two, and C at zero, zero.



What is the length of segment


?
units

1 answer

To find the length of segment \( AB' \) in the triangle with vertices \( A' \), \( B \), and \( C \), we first need the coordinates of the points involved.

From your description:

  • The vertex \( A' \) is located at \( (0, 10) \).
  • The vertex \( B \) is at \( (4, 2) \).
  • The vertex \( C \) appears to be at the origin \( (0, 0) \).

Now, we can calculate the length of segment \( AB' \) between points \( A' \) and \( B \).

Using the distance formula, which is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

we can substitute the coordinates of \( A' \) and \( B \):

  • \( A' = (0, 10) \)
  • \( B = (4, 2) \)

Now, substituting the coordinates into the distance formula:

\[ AB' = \sqrt{(4 - 0)^2 + (2 - 10)^2} \] \[ = \sqrt{(4)^2 + (-8)^2} \] \[ = \sqrt{16 + 64} \] \[ = \sqrt{80} \] \[ = \sqrt{16 \times 5} \] \[ = 4\sqrt{5} \]

Therefore, the length of segment \( AB' \) is:

\[ 4\sqrt{5} \text{ units} \]