A coordinate plane's axes range from negative 3 to 7 by 1-unit increments. 3 points are plotted and joined by solid lines. The points are labeled upper A, upper B, and upper C.
AC¯¯¯¯¯¯¯¯
is a dilation of AB¯¯¯¯¯¯¯¯
, with point B
as the center of dilation. Determine the scale factor.
To find the scale factor, we need to compare the lengths of AC and AB.
To find the length of AC, we can use the distance formula. Let the coordinates of point A be (x1, y1) and the coordinates of point C be (x2, y2).
From the given coordinate range, it appears that the x-coordinates range from -3 to 7, and the y-coordinates also range from -3 to 7.
Using the distance formula, we have:
AC = √((x2 - x1)^2 + (y2 - y1)^2)
Since AC is a dilation of AB, the coordinates of points A, B, and C are related by a scale factor. Let the scale factor be k.
Therefore, the coordinates of point C, (x2, y2), can be found by multiplying the coordinates of point B, (x1, y1), by the scale factor:
x2 = k * x1
y2 = k * y1
Substituting these values into the distance formula, we have:
AC = √((k * x1 - x1)^2 + (k * y1 - y1)^2)
= √((k - 1)^2 * (x1^2 + y1^2))
= √((k - 1)^2) * √(x1^2 + y1^2)
= |k - 1| * √(x1^2 + y1^2)
Since we know that AC¯¯¯¯¯¯¯¯ is a dilation of AB¯¯¯¯¯¯¯¯, we can assume that AC is longer than AB. Therefore, |k - 1| > 1.
To find the scale factor, we need to find the value of k that makes AC¯¯¯¯¯¯¯¯ equal to AB¯¯¯¯¯¯¯¯. This means that the length of AC must be the same as the length of AB.
Thus, we have:
|k - 1| * √(x1^2 + y1^2) = √(x1^2 + y1^2)
The square roots on both sides of the equation can be canceled out:
|k - 1| = 1
This equation has two possible solutions:
1) k - 1 = 1
k = 2
2) k - 1 = -1
k = 0
However, since |k - 1| must be greater than 1, the scale factor cannot be 0. Therefore, the scale factor is k = 2.
Therefore, AC¯¯¯¯¯¯¯¯ is a dilation of AB¯¯¯¯¯¯¯¯ with a scale factor of 2.