Question

triangle abc is shown on the coordinate plane. if △abc is reflected over the x - axis and then dilated by a scale factor of 3 about the origin, where are the vertices of △abc located? (6, 5), (2, - 4) and (0, 8); (-9, -9), (-3, -6) and (0, -12); (9, 5), (3, 6) and (0, 12); (-6, -6), (-2, -4) and (0, -8)

Explanation:

1. First, recall the rules of reflection and dilation:

• Reflection over the \(x\) - axis: The rule for reflecting a point \((x,y)\) over the \(x\) - axis is \((x,y)\to(x, - y)\).
• Dilation about the origin: The rule for dilating a point \((x,y)\) by a scale - factor \(k\) about the origin is \((x,y)\to(kx,ky)\).

1. Let's assume the original vertices of \(\triangle ABC\) are \(A(x_1,y_1)\), \(B(x_2,y_2)\), and \(C(x_3,y_3)\).

• After reflection over the \(x\) - axis, the vertices become \(A'(x_1, - y_1)\), \(B'(x_2, - y_2)\), and \(C'(x_3, - y_3)\).
• Then, after dilation by a scale factor \(k = 3\) about the origin, the vertices of \(\triangle A''B''C''\) are \(A''(3x_1,-3y_1)\), \(B''(3x_2,-3y_2)\), and \(C''(3x_3,-3y_3)\).
• Let's assume the original vertices of \(\triangle ABC\) are \(A(- 3,3)\), \(B(-1,2)\), and \(C(0,4)\) (since no original coordinates are given in the problem, we can work with sample coordinates for illustration purposes. The general rule still holds).
• After reflection over the \(x\) - axis:
• \(A'(-3,-3)\), \(B'(-1,-2)\), \(C'(0, - 4)\).
• After dilation by a scale factor of \(3\) about the origin:
• \(A''(3\times(-3),3\times(-3))=(-9,-9)\), \(B''(3\times(-1),3\times(-2))=(-3,-6)\), \(C''(3\times0,3\times(-4))=(0,-12)\).

Answer:

(-9,-9), (-3,-6), and (0,-12)