Triangle XYZ is drawn on a coordinate plane with vertices X(0,4), Y(2,0), and Z(4,7). If you dilate the triangle to form triangle X′Y′Z′ using a magnitude of 1/2 and a center of dilation at (4,7), what are the coordinate points of the new vertices?(1 point)

Question

Responses

X′(−4,−3), Y′(−2,−7), Z′(0,0)
upper X prime left parenthesis negative 4 comma negative 3 right parenthesis , upper Y prime left parenthesis negative 2 comma negative 7 right parenthesis , upper Z prime left parenthesis 0 comma 0 right parenthesis

X′(2,5.5), Y′(3,3.5), Z′(4,7)
upper X prime left parenthesis 2 comma 5.5 right parenthesis , upper Y prime left parenthesis 3 comma 3.5 right parenthesis , upper Z prime left parenthesis 4 comma 7 right parenthesis

X′(4,11), Y′(6,7), Z′(8,14)
upper X prime left parenthesis negative 4 comma 11 right parenthesis , upper Y prime left parenthesis 6 comma 7 right parenthesis , upper Z prime left parenthesis 8 comma 14 right parenthesis

X′(−4,1), Y′(0,−7), Z′(4,7)

Answer 1

To find the coordinates of the new vertices \(X′Y′Z′\) after dilating triangle \(XYZ\) with a center of dilation at \(C(4,7)\) and a magnitude of \( \frac{1}{2} \), we can use the formula for dilation.

The formula for dilation of a point \( (x, y) \) from a center of dilation \( (x_c, y_c) \) is given by: \[ (x', y') = (x_c + k(x - x_c), y_c + k(y - y_c)) \] where \( k \) is the scale factor.

Now, let’s find the new coordinates of each vertex:

1. Vertex X(0, 4): \[ X' = (4 + \frac{1}{2}(0 - 4), 7 + \frac{1}{2}(4 - 7)) \] \[ X' = (4 - 2, 7 - 1.5) = (2, 5.5) \]

2. Vertex Y(2, 0): \[ Y' = (4 + \frac{1}{2}(2 - 4), 7 + \frac{1}{2}(0 - 7)) \] \[ Y' = (4 - 1, 7 - 3.5) = (3, 3.5) \]

3. Vertex Z(4, 7): \[ Z' = (4 + \frac{1}{2}(4 - 4), 7 + \frac{1}{2}(7 - 7)) \] \[ Z' = (4 + 0, 7 + 0) = (4, 7) \]

Thus, the new vertices after dilation are:

• \(X′(2, 5.5)\)
• \(Y′(3, 3.5)\)
• \(Z′(4, 7)\)

The correct response from the options provided is: X′(2,5.5), Y′(3,3.5), Z′(4,7).