Question

dilate △abc on the coordinate plane using the origin (0, 0) as the center of dilation and a scale factor of $\frac{1}{2}$.

Explanation:

1. First, identify the coordinates of the vertices of \(\triangle ABC\):

• Assume the coordinates of point \(A\) are \((x_A,y_A)\), point \(B\) are \((x_B,y_B)\), and point \(C\) are \((x_C,y_C)\) from the graph. Let's say \(A=(8,10)\), \(B=(10,18)\), \(C=(16,10)\).
• The formula for dilation with the origin \((0,0)\) as the center of - dilation and a scale - factor \(k\) is \((x',y') = k(x,y)\), where \((x,y)\) are the original coordinates and \((x',y')\) are the coordinates of the dilated point. Here \(k = \frac{1}{2}\).

1. Calculate the coordinates of the dilated points:

• For point \(A\):
• \(x_{A}'=\frac{1}{2}x_A\) and \(y_{A}'=\frac{1}{2}y_A\). Substituting \(x_A = 8\) and \(y_A = 10\), we get \(x_{A}'=\frac{1}{2}\times8 = 4\) and \(y_{A}'=\frac{1}{2}\times10 = 5\). So the new coordinates of \(A'\) are \((4,5)\).
• For point \(B\):
• \(x_{B}'=\frac{1}{2}x_B\) and \(y_{B}'=\frac{1}{2}y_B\). Substituting \(x_B = 10\) and \(y_B = 18\), we get \(x_{B}'=\frac{1}{2}\times10 = 5\) and \(y_{B}'=\frac{1}{2}\times18 = 9\). So the new coordinates of \(B'\) are \((5,9)\).
• For point \(C\):
• \(x_{C}'=\frac{1}{2}x_C\) and \(y_{C}'=\frac{1}{2}y_C\). Substituting \(x_C = 16\) and \(y_C = 10\), we get \(x_{C}'=\frac{1}{2}\times16 = 8\) and \(y_{C}'=\frac{1}{2}\times10 = 5\). So the new coordinates of \(C'\) are \((8,5)\).

The vertices of the dilated triangle \(\triangle A'B'C'\) are \(A'(4,5)\), \(B'(5,9)\), \(C'(8,5)\).

Answer:

1. First, identify the coordinates of the vertices of \(\triangle ABC\):

• Assume the coordinates of point \(A\) are \((x_A,y_A)\), point \(B\) are \((x_B,y_B)\), and point \(C\) are \((x_C,y_C)\) from the graph. Let's say \(A=(8,10)\), \(B=(10,18)\), \(C=(16,10)\).
• The formula for dilation with the origin \((0,0)\) as the center of - dilation and a scale - factor \(k\) is \((x',y') = k(x,y)\), where \((x,y)\) are the original coordinates and \((x',y')\) are the coordinates of the dilated point. Here \(k = \frac{1}{2}\).

1. Calculate the coordinates of the dilated points:

• For point \(A\):
• \(x_{A}'=\frac{1}{2}x_A\) and \(y_{A}'=\frac{1}{2}y_A\). Substituting \(x_A = 8\) and \(y_A = 10\), we get \(x_{A}'=\frac{1}{2}\times8 = 4\) and \(y_{A}'=\frac{1}{2}\times10 = 5\). So the new coordinates of \(A'\) are \((4,5)\).
• For point \(B\):
• \(x_{B}'=\frac{1}{2}x_B\) and \(y_{B}'=\frac{1}{2}y_B\). Substituting \(x_B = 10\) and \(y_B = 18\), we get \(x_{B}'=\frac{1}{2}\times10 = 5\) and \(y_{B}'=\frac{1}{2}\times18 = 9\). So the new coordinates of \(B'\) are \((5,9)\).
• For point \(C\):
• \(x_{C}'=\frac{1}{2}x_C\) and \(y_{C}'=\frac{1}{2}y_C\). Substituting \(x_C = 16\) and \(y_C = 10\), we get \(x_{C}'=\frac{1}{2}\times16 = 8\) and \(y_{C}'=\frac{1}{2}\times10 = 5\). So the new coordinates of \(C'\) are \((8,5)\).

The vertices of the dilated triangle \(\triangle A'B'C'\) are \(A'(4,5)\), \(B'(5,9)\), \(C'(8,5)\).