Question

find point ( h ) to prove that ( overline{dh} ) is an enlargement of ( overline{df} ) by a scale factor of 2 with point ( d ) as the center of dilation. (1 point) ( \bigcirc h ) is at ( (5.5,3) ) ( \bigcirc h ) is at ( (8,18) ) ( \bigcirc h ) is at ( (13,6) ) ( \bigcirc h ) is at ( (-2,0) )

Explanation:

Step1: Find the coordinates of point \(D\) and \(F\)

From the graph, \(D=(2,2)\) and \(F=(7,4)\)

Step2: Use the dilation formula

If the center of dilation is \(C=(x_0,y_0)\) and the scale factor is \(k\), and a point \(P=(x,y)\) is dilated to \(P'=(x',y')\), then the formula is \(x'=x_0 + k(x - x_0)\) and \(y'=y_0 + k(y - y_0)\)
Here \(k = 2\), \(x_0=2\), \(y_0 = 2\), \(x = 7\), \(y=4\)
For \(x\)-coordinate: \(x'=2+2\times(7 - 2)=2 + 10=12\)
For \(y\)-coordinate: \(y'=2+2\times(4 - 2)=2+4 = 6\)
Another way: The vector from \(D\) to \(F\) is \(\overrightarrow{DF}=(7 - 2,4 - 2)=(5,2)\)
After dilation with scale factor \(k = 2\) from center \(D\), the vector \(\overrightarrow{DH}=2\overrightarrow{DF}=(10,4)\)
So \(H=(2+10,2 + 4)=(12,6)\) (There might be a graph - reading error. If we assume a unit - based approach:
The horizontal distance from \(D\) to \(F\): \(7-2 = 5\) units. After dilation with scale factor \(2\), the horizontal distance from \(D\) to \(H\) is \(10\) units. So \(x\) - coordinate of \(H\) is \(2+10=12\)
The vertical distance from \(D\) to \(F\): \(4 - 2=2\) units. After dilation with scale factor \(2\), the vertical distance from \(D\) to \(H\) is \(4\) units. So \(y\) - coordinate of \(H\) is \(2+4 = 6\). Maybe due to graph grid mis - reading (if we consider a wrong count of units, assume each grid is \(1\) unit. If we check the options:
Let's use the formula \(x=x_D+k(x_F - x_D)\), \(y=y_D+k(y_F - y_D)\)
If \(x_D = 2\), \(y_D=2\), \(x_F\) (assume mis - read as \(3 + 5=8\) in some wrong grid - counting, \(y_F\) (assume mis - read as \(2+2 = 4\))
\(x=2+2\times(8 - 2)=14\) (wrong), \(y=2+2\times(4 - 2)=6\)
If we consider the general dilation formula for a point \(P(x,y)\) dilated about \(C(x_c,y_c)\) as \(P'(x_c + k(x - x_c),y_c + k(y - y_c))\)
For option \(H=(13,6)\):
Let \(x_c = 2\), \(y_c=2\), \(k = 2\)
\(x=2+2\times(x_F-2)\), \(6=2+2\times(y_F - 2)\) gives \(y_F=4\)
\(13=2+2\times(x_F - 2)\) gives \(x_F=\frac{13 - 2}{2}+2=\frac{11}{2}+2=7.5\) (approximate graph - reading where \(F\) is at \((8,4)\) (if we assume a half - unit mis - count in \(x\) - direction of \(F\))

Answer:

\(H\) is at \((13,6)\)