Question

triangle xyz is rotated 270° counterclockwise about the origin. the result is △xyz, as shown below. (a) the arrows below show that the coordinates on the left are mapped to the coordinates on the right. fill in the blanks to give the coordinates after the rotation. original coordinates → final coordinates x(-8,1)→x(□,□) y(-6,7)→y(□,□) z(-2,-3)→z(□,□) (b) choose the general rule below that describes the rotation mapping △xyz to △xyz. (x,y)→(-x,y) (x,y)→(x,-y) (x,y)→(y,x) (x,y)→(-y,x) (x,y)→(-x,-y) (x,y)→(y,-x) (x,y)→(-y,-x)

Explanation:

Step1: Recall the rotation rule

When a point \((x,y)\) is rotated \(270^{\circ}\) counter - clockwise about the origin, the transformation rule is \((x,y)\to(y, - x)\).

Step2: Apply the rule to point \(X(-8,1)\)

For \(X(-8,1)\), substitute \(x=-8\) and \(y = 1\) into the rule \((x,y)\to(y,-x)\).
We get \(X'(1,8)\).

Step3: Apply the rule to point \(Y(-6,7)\)

For \(Y(-6,7)\), substitute \(x=-6\) and \(y = 7\) into the rule \((x,y)\to(y,-x)\).
We get \(Y'(7,6)\).

Step4: Apply the rule to point \(Z(-2,-3)\)

For \(Z(-2,-3)\), substitute \(x=-2\) and \(y=-3\) into the rule \((x,y)\to(y,-x)\).
We get \(Z'(-3,2)\).

Step5: Verify the rule for part (b)

The rule \((x,y)\to(y,-x)\) is one of the options.

Answer:

(a) \(X(-8,1)\to X'(1,8)\); \(Y(-6,7)\to Y'(7,6)\); \(Z(-2,-3)\to Z'(-3,2)\)
(b) \((x,y)\to(y,-x)\)