QUESTION IMAGE
Question
angle def is rotated 90° counterclockwise about the origin.
the result is △def, 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
d(-8, -2) → d(□,□)
e(1, -4) → e(□,□)
f(-2, -6) → f(□,□)
(b) choose the general rule below that describes the rotation
mapping △def to △def.
○(x,y)→(x, -y) ○(x,y)→(-x,y)
○(x,y)→(-x, -y) ○(x,y)→(-y,x)
○(x,y)→(-y, -x) ○(x,y)→(y,x)
○(x,y)→(y, -x)
Step1: Recall the rule for 90° counter - clockwise rotation
The rule for a 90° counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\).
Step2: Apply the rule to point \(D(-8,-2)\)
For \(D(-8,-2)\), substitute \(x = - 8\) and \(y=-2\) into the rule \((x,y)\to(-y,x)\).
We get \(-y=-(-2) = 2\) and \(x=-8\), so \(D'=(2,-8)\).
Step3: Apply the rule to point \(E(1,-4)\)
For \(E(1,-4)\), substitute \(x = 1\) and \(y = - 4\) into the rule \((x,y)\to(-y,x)\).
We get \(-y=-(-4)=4\) and \(x = 1\), so \(E'=(4,1)\).
Step4: Apply the rule to point \(F(-2,-6)\)
For \(F(-2,-6)\), substitute \(x=-2\) and \(y=-6\) into the rule \((x,y)\to(-y,x)\).
We get \(-y=-(-6)=6\) and \(x=-2\), so \(F'=(6,-2)\).
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(a) \(D(-8,-2)\to D'(2,-8)\), \(E(1,-4)\to E'(4,1)\), \(F(-2,-6)\to F'(6,-2)\)
(b) \((x,y)\to(-y,x)\)