Question

graph the image of \\( \triangle def \\) after a reflection over the line \\( x = 3 \\).

Explanation:

Step1: Find coordinates of \(\triangle DEF\)

Assume \(D(-1,-8)\), \(E(-1,-7)\), \(F(-4,-6)\)

Step2: Use reflection formula

The formula for reflection over the line \(x = a\) is \((x,y)\to(2a - x,y)\). Here \(a = 3\), so \((x,y)\to(6 - x,y)\)

• For \(D(-1,-8)\): \(x=-1,y = -8\), new \(x=6-(-1)=7\), so \(D'(7,-8)\)
• For \(E(-1,-7)\): \(x=-1,y=-7\), new \(x=6-(-1)=7\), so \(E'(7,-7)\)
• For \(F(-4,-6)\): \(x = -4,y=-6\), new \(x=6-(-4)=10\), so \(F'(10,-6)\)

Step3: Plot the new points

Plot \(D'(7,-8)\), \(E'(7,-7)\) and \(F'(10,-6)\) on the coordinate plane and connect them to form \(\triangle D'E'F'\)

Answer:

Plot the points \(D'(7,-8)\), \(E'(7,-7)\) and \(F'(10,-6)\) and connect them to get the reflected triangle.