Question

write the coordinates of the vertices after a reflection over the line ( y = -1 ).
( t(square,square) )
( u(square,square) )
( v(square,square) )
( w(square,square) )

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 \(d=\vert y - (-1)\vert=\vert y + 1\vert\).

• For \(T(6,-9)\): \(d_T=\vert-9+1\vert = 8\)
• For \(U(8,-7)\): \(d_U=\vert-7 + 1\vert=6\)
• For \(V(6,-5)\): \(d_V=\vert-5+1\vert = 4\)
• For \(W(-1,-7)\): \(d_W=\vert-7+1\vert=6\)

Step2: Calculate the new \(y\) - coordinate

The formula for reflecting a point \((x,y)\) over the line \(y = k\) is \((x,2k - y)\). Here \(k=-1\), so the new \(y\) - coordinate is \(y'=2\times(-1)-y=-2 - y\), and the \(x\) - coordinate remains the same.

• For \(T(6,-9)\): \(x = 6\), \(y'=-2-(-9)=7\), so \(T'(6,7)\)
• For \(U(8,-7)\): \(x = 8\), \(y'=-2-(-7)=5\), so \(U'(8,5)\)
• For \(V(6,-5)\): \(x = 6\), \(y'=-2-(-5)=3\), so \(V'(6,3)\)
• For \(W(-1,-7)\): \(x=-1\), \(y'=-2-(-7)=5\), so \(W'(-1,5)\)

Answer:

\(T'(6,7)\), \(U'(8,5)\), \(V'(6,3)\), \(W'(-1,5)\)