Question

what would the center of dilation be for the following object with d at the point (-6,3) and f at the point (-6,1.5)? (-7,2) (-7,3) (-8,2) (-8,3)

Explanation:

Step1: Recall dilation formula

For a dilation with center \((a,b)\), the formula for a point \((x,y)\) is \((x',y')=(a + k(x - a),b + k(y - b))\). Here we are not given the scale - factor \(k\) explicitly, but we can use the fact that we know the center of dilation \((-6,1.5)\), the image of one point \(D'\) at \((-6,3)\) and we want to find the pre - image of \(F'\) at \((-6,1.5)\).

Step2: Analyze vertical movement

The \(x\) - coordinate of the center of dilation is \(-6\). The \(x\) - coordinate of \(D'\) is \(-6\) and we know that for a point \((x,y)\) dilated about \((a,b)\) with \(a=-6\), the \(x\) - coordinate of the image \(x'=-6 + k(x + 6)\). Since \(x'\) of \(D'\) is \(-6\), it means the horizontal position relative to the center of dilation for points in this problem behaves in a certain way. For the \(y\) - coordinate, the center of dilation is \(y = 1.5\), the \(y\) - coordinate of \(D'\) is \(y_{D'}=3\). The distance from the center of dilation to \(D'\) in the \(y\) - direction is \(3-1.5 = 1.5\).
The \(y\) - coordinate of \(F'\) is \(1.5\). If we assume a linear relationship based on the dilation about \((-6,1.5)\), and knowing the position of \(D'\), we can work out the pre - image of \(F'\). By observing the relationship between the center of dilation \((-6,1.5)\) and \(D'\) at \((-6,3)\) and applying the same dilation logic to \(F'\), we find that the pre - image of \(F'\) has coordinates \((-7,2)\).

Answer:

(-7,2)