Question

graph the image of \\( \triangle efg \\) after a reflection over the line \\( y = 1 \\).

Explanation:

Step1: Find the distance from each point to the line \(y = 1\)

For a point \((x,y)\), the distance to the line \(y = 1\) is \(d=\vert y - 1\vert\).

• For point \(E(0,5)\), \(d_E=\vert5 - 1\vert=4\)
• For point \(F(1,5)\), \(d_F=\vert5 - 1\vert=4\)
• For point \(G(0,3)\), \(d_G=\vert3 - 1\vert=2\)

Step2: Reflect each point over the line \(y = 1\)

The formula for reflecting a point \((x,y)\) over the line \(y = k\) is \((x,2k - y)\). Here \(k = 1\), so the formula is \((x,2\times1 - y)=(x,2 - y)\)

• For \(E(0,5)\): \(E'(0,2-5)=(0,-3)\)
• For \(F(1,5)\): \(F'(1,2 - 5)=(1,-3)\)
• For \(G(0,3)\): \(G'(0,2 - 3)=(0,-1)\)

Answer:

Plot the points \(E'(0,-3)\), \(F'(1,-3)\) and \(G'(0,-1)\) and connect them to form the reflected triangle \(\triangle E'F'G'\)