triangle xyz has vertices x(1, 3), y(0, 0), and z(-1, 2). the image of triangle xyz after a rotation has…

triangle xyz has vertices x(1, 3), y(0, 0), and z(-1, 2). the image of triangle xyz after a rotation has vertices x(-3, 1), y(0, 0), and z(-2, -1). which rule describes the transformation?\no $r_{0,90^{circ}}$\no $r_{0,180^{circ}}$\no $r_{0,270^{circ}}$\no $r_{0,360^{circ}}$

triangle xyz has vertices x(1, 3), y(0, 0), and z(-1, 2). the image of triangle xyz after a rotation has vertices x(-3, 1), y(0, 0), and z(-2, -1). which rule describes the transformation?\no $r_{0,90^{circ}}$\no $r_{0,180^{circ}}$\no $r_{0,270^{circ}}$\no $r_{0,360^{circ}}$

Answer

Explanation:

Step1: Recall rotation rules

The general rule for rotating a point $(x,y)$ counter - clockwise about the origin $O(0,0)$ by an angle $\theta$ is: For a $90^{\circ}$ rotation: $(x,y)\to(-y,x)$; for a $180^{\circ}$ rotation: $(x,y)\to(-x,-y)$; for a $270^{\circ}$ rotation: $(x,y)\to(y, - x)$; for a $360^{\circ}$ rotation: $(x,y)\to(x,y)$.

Step2: Check the rotation of point $X(1,3)$

If we rotate point $X(1,3)$ by $270^{\circ}$ counter - clockwise about the origin, using the rule $(x,y)\to(y,-x)$, we substitute $x = 1$ and $y = 3$ to get $(3,-1)$. This is incorrect. If we rotate point $X(1,3)$ by $180^{\circ}$ counter - clockwise about the origin, using the rule $(x,y)\to(-x,-y)$, we substitute $x = 1$ and $y = 3$ to get $(-1,-3)$. This is incorrect. If we rotate point $X(1,3)$ by $90^{\circ}$ counter - clockwise about the origin, using the rule $(x,y)\to(-y,x)$, we substitute $x = 1$ and $y = 3$ to get $(-3,1)$ which is the coordinates of $X'$.

Step3: Check other points

For point $Y(0,0)$, rotating it by $90^{\circ}$ counter - clockwise about the origin using the rule $(x,y)\to(-y,x)$ gives $(0,0)$ which is $Y'$. For point $Z(-1,2)$, rotating it by $90^{\circ}$ counter - clockwise about the origin using the rule $(x,y)\to(-y,x)$ gives $(-2,-1)$ which is $Z'$.

Answer:

$R_{0,90^{\circ}}$