Question

horizontal lines

1. triangle: a(-4, 1), b(-1, 1), and c(-3, 4)

reflection: reflection over the line y = -3
a (____, ____)
b (____, ____)
c (____, ____)
rule: think midpoint
a(-4, 1), b(-1, 1), and c(-3, 4)
( , ) ( , ) ( , )
a( , ), b( , ) c( , )

1. triangle: a(2, 3), b(5, 6), and c(8, 4)

reflection: reflection over the line x = -1
a (____, ____)
b (____, ____)
c (____, ____)
rule: think midpoint
a(2, 3), b(5, 6), and c(8, 4)
( , ) ( , ) ( , )
a( , ), b( , ) c( , )

Explanation:

Step1: Reflection over \(y = - 3\)

For a point \((x,y)\) reflected over the line \(y = k\), the formula is \((x,2k - y)\). Here \(k=-3\).
For \(A(-4,1)\):
\(x=-4\), \(y = 1\), \(2k - y=2\times(-3)-1=-6 - 1=-7\). So \(A'(-4,-7)\)
For \(B(-1,1)\):
\(x=-1\), \(y = 1\), \(2k - y=2\times(-3)-1=-7\). So \(B'(-1,-7)\)
For \(C(-3,4)\):
\(x=-3\), \(y = 4\), \(2k - y=2\times(-3)-4=-6 - 4=-10\). So \(C'(-3,-10)\)

Step2: Reflection over \(x=-1\)

For a point \((x,y)\) reflected over the line \(x = h\), the formula is \((2h - x,y)\). Here \(h = - 1\)
For \(A(2,3)\):
\(x = 2\), \(y=3\), \(2h - x=2\times(-1)-2=-2 - 2=-4\). So \(A'(-4,3)\)
For \(B(5,6)\):
\(x = 5\), \(y = 6\), \(2h - x=2\times(-1)-5=-2 - 5=-7\). So \(B'(-7,6)\)
For \(C(8,4)\):
\(x = 8\), \(y = 4\), \(2h - x=2\times(-1)-8=-2 - 8=-10\). So \(C'(-10,4)\)

Answer:

For reflection over \(y=-3\):
\(A'(-4,-7)\), \(B'(-1,-7)\), \(C'(-3,-10)\)
For reflection over \(x = - 1\):
\(A'(-4,3)\), \(B'(-7,6)\), \(C'(-10,4)\)