Question

1. parallelogram qrst with vertices ( q(2,-3) ), ( r(7,1), s(6,-2) ), and ( t(1,-6) ) (a) reflection: in the line ( y = x ) (b) translation: ( (x,y)\to(x - 3,y - 7) )

Explanation:

Step1: Reflection in the line \(y = x\)

The rule for reflection in the line \(y=x\) is \((x,y)\to(y,x)\).
For point \(Q(2,-3)\), after reflection \(Q'( - 3,2)\).
For point \(R(7,1)\), after reflection \(R'(1,7)\).
For point \(S(6,-2)\), after reflection \(S'(-2,6)\).
For point \(T(1,-4)\), after reflection \(T'(-4,1)\).

Step2: Translation \((x,y)\to(x - 3,y-7)\)

The rule for translation \((x,y)\to(x - 3,y - 7)\) means we subtract \(3\) from the \(x\)-coordinate and \(7\) from the \(y\)-coordinate.
For \(Q'(-3,2)\): \(x=-3-3=-6\), \(y = 2-7=-5\), so \(Q''(-6,-5)\).
For \(R'(1,7)\): \(x=1 - 3=-2\), \(y=7-7 = 0\), so \(R''(-2,0)\).
For \(S'(-2,6)\): \(x=-2-3=-5\), \(y=6 - 7=-1\), so \(S''(-5,-1)\).
For \(T'(-4,1)\): \(x=-4-3=-7\), \(y=1-7=-6\), so \(T''(-7,-6)\).

Answer:

\(Q'(-6,-5)\), \(R'(-2,0)\), \(S'(-5,-1)\), \(T'(-7,-6)\)