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rotating $\\triangle abk$ by $90^\\circ$ counterclockwise about the ori…

Question

rotating $\triangle abk$ by $90^\circ$ counterclockwise about the origin, we get its image $\triangle abk$ as shown below.
note that $\triangle abk$ has vertices $a(-1, -6)$, $b(-5, -5)$, and $k(-7, -2)$.
also, note that $\triangle abk$ has vertices $a(6, -1)$, $b(5, -5)$, and $k(2, -7)$.
complete the following.

scratch area (not part of answer)

(a) suppose each pair of segments below have the same length.
find each length. give exact answers (not decimal approximations).
$ca = ca = \square$ units

$cb = cb = \square$ units

$ck = ck = \square$ units

(b) suppose the angles below have the same measure. choose the correct angle
measure. use the protractor provided, as necessary.
$m\angle aca = m\angle bcb = m\angle kck = \text{select}$

(c) choose the correct pair of statements about the rotation.

  • not every point on the original figure is the same distance from the

center of rotation as its image.

  • not all angles formed by a point and its image, with the vertex at the

center of rotation, are congruent.

  • each point on the original figure is the same distance from the center of

rotation as its image.

  • not all angles formed by a point and its image, with the vertex at the

center of rotation, are congruent.

  • not every point on the original figure is the same distance from the

center of rotation as its image.

  • all angles formed by a point and its image, with the vertex at the center

of rotation, are congruent.

  • each point on the original figure is the same distance from the center of

rotation as its image.

  • all angles formed by a point and its image, with the vertex at the center

of rotation, are congruent.

Explanation:

Step1: Identify center C

From the grid, $C$ is the origin $(0,0)$.

Step2: Calculate $CA$ (distance formula)

Distance formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$CA = \sqrt{(-1-0)^2+(-6-0)^2} = \sqrt{1+36} = \sqrt{37}$

Step3: Calculate $CB$ (distance formula)

$CB = \sqrt{(-5-0)^2+(-5-0)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}$

Step4: Calculate $CK$ (distance formula)

$CK = \sqrt{(-7-0)^2+(-2-0)^2} = \sqrt{49+4} = \sqrt{53}$

Step5: Determine rotation angle

The figure is rotated $90^\circ$ counterclockwise, so the central angles equal the rotation angle.

Step6: Analyze rotation properties

Rotations preserve distance from center, and all central angles for a rotation are congruent.

Answer:

(a)

$CA = CA' = \sqrt{37}$ units
$CB = CB' = 5\sqrt{2}$ units
$CK = CK' = \sqrt{53}$ units

(b)

$m\angle ACA' = m\angle BCB' = m\angle KCK' = 90^\circ$

(c)

Each point on the original figure is the same distance from the center of rotation as its image.
All angles formed by a point and its image, with the vertex at the center of rotation, are congruent.