Question

select all of the transformations that preserve distance.

$(x,y)\to(y - 4,x + 3)$

$(x,y)\to(y,-x)$

$(x,y)\to(0.5x,-0.5x)$

$(x,y)\to(x,y)$

$(x,y)\to(-x - 2,y + 9)$

$(x,y)\to(7x,7y)$

Explanation:

Step1: Recall the definition of distance - preserving transformations

Isometries (distance - preserving transformations) include translations, rotations, and reflections. A translation has the form \((x,y)\to(x + a,y + b)\), a rotation (about the origin) can be represented by a linear transformation, and a reflection also has a specific linear form.

Step2: Analyze each transformation

• For \((x,y)\to(y - 4,x + 3)\):

Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\). The distance between \(P\) and \(Q\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). After transformation, \(P'\) has coordinates \((y_1 - 4,x_1+3)\) and \(Q'\) has coordinates \((y_2 - 4,x_2 + 3)\). The distance \(d'=\sqrt{(y_2 - y_1)^2+(x_2 - x_1)^2}=d\). This is a combination of a reflection (swap \(x\) and \(y\)) and a translation.

• For \((x,y)\to(y,-x)\):

Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\). The distance between \(P\) and \(Q\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). After transformation, \(P'\) has coordinates \((y_1,-x_1)\) and \(Q'\) has coordinates \((y_2,-x_2)\). The distance \(d'=\sqrt{(y_2 - y_1)^2+(-x_2+x_1)^2}=\sqrt{(y_2 - y_1)^2+(x_2 - x_1)^2}=d\). This is a rotation of \(270^{\circ}\) about the origin.

• For \((x,y)\to(0.5x,-0.5x)\):

Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\). The distance between \(P\) and \(Q\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). After transformation, \(P'\) has coordinates \((0.5x_1,-0.5x_1)\) and \(Q'\) has coordinates \((0.5x_2,-0.5x_2)\). The distance \(d'=\sqrt{(0.5x_2 - 0.5x_1)^2+(-0.5x_2 + 0.5x_1)^2}=\frac{\sqrt{2}}{2}\vert x_2 - x_1\vert
eq d\) (unless \(y_1=y_2 = 0\) and \(x_2=x_1\)).

• For \((x,y)\to(x,y)\):

This is the identity transformation. Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\). The distance between \(P\) and \(Q\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). After transformation, \(P'\) has coordinates \((x_1,y_1)\) and \(Q'\) has coordinates \((x_2,y_2)\). So \(d' = d\).

• For \((x,y)\to(-x - 2,y + 9)\):

Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\). The distance between \(P\) and \(Q\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). After transformation, \(P'\) has coordinates \((-x_1 - 2,y_1+9)\) and \(Q'\) has coordinates \((-x_2 - 2,y_2 + 9)\). The distance \(d'=\sqrt{(-x_2+x_1)^2+(y_2 - y_1)^2}=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}=d\). This is a combination of a reflection (over the \(y\) - axis) and a translation.

• For \((x,y)\to(7x,7y)\):

Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\). The distance between \(P\) and \(Q\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). After transformation, \(P'\) has coordinates \((7x_1,7y_1)\) and \(Q'\) has coordinates \((7x_2,7y_2)\). The distance \(d'=\sqrt{(7x_2 - 7x_1)^2+(7y_2 - 7y_1)^2}=7\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}
eq d\) (unless \(x_2=x_1\) and \(y_2=y_1\)).

Answer:

\((x,y)\to(y - 4,x + 3)\), \((x,y)\to(y,-x)\), \((x,y)\to(x,y)\), \((x,y)\to(-x - 2,y + 9)\)