Question

write the coordinates of the vertices after a reflection over the line y = 2.

Explanation:

1. Recall the rule for reflecting a point \((x,y)\) over the horizontal line \(y = k\):

• The formula for reflecting a point \((x,y)\) over the horizontal line \(y = k\) is \((x,2k - y)\). Here \(k = 2\), so the rule is \((x,4 - y)\).

1. Assume the coordinates of the vertices are:

• Let's assume \(T(x_T,y_T)\), \(U(x_U,y_U)\), \(V(x_V,y_V)\), and \(W(x_W,y_W)\). From the graph, if we assume \(T(- 6,1)\), \(U(-4,2)\), \(V(-6,4)\), \(W(-8,2)\).
• For point \(T(-6,1)\):
• Using the formula \((x,4 - y)\), we substitute \(x=-6\) and \(y = 1\). Then the new - \(y\) value is \(4-1 = 3\), and the \(x\) - value remains the same. So the new coordinates of \(T\) are \(T'(-6,3)\).
• For point \(U(-4,2)\):
• Substitute \(x=-4\) and \(y = 2\) into the formula \((x,4 - y)\). We get \(4 - 2=2\), and the \(x\) - value remains \(-4\). So the new coordinates of \(U\) are \(U'(-4,2)\).
• For point \(V(-6,4)\):
• Substitute \(x=-6\) and \(y = 4\) into the formula \((x,4 - y)\). We have \(4 - 4 = 0\), and the \(x\) - value remains \(-6\). So the new coordinates of \(V\) are \(V'(-6,0)\).
• For point \(W(-8,2)\):
• Substitute \(x=-8\) and \(y = 2\) into the formula \((x,4 - y)\). We get \(4 - 2 = 2\), and the \(x\) - value remains \(-8\). So the new coordinates of \(W\) are \(W'(-8,2)\).

Answer:

The coordinates of the vertices after reflection over the line \(y = 2\) are \(T'(-6,3)\), \(U'(-4,2)\), \(V'(-6,0)\), \(W'(-8,2)\)