Question

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rotating \\( \triangle j k l \\) by \\( 180^{circ} \\) counterclockwise about the origin, we get its image \\( \triangle j^{prime} k^{prime} l^{prime} \\) as shown below. note that \\( \triangle j k l \\) has vertices \\( j(-3,7), k(-5,3) \\), and \\( l(-8,2) \\). also, note that \\( \triangle j^{prime} k^{prime} l^{prime} \\) has vertices \\( j^{prime}(3,-7), k^{prime}(5,-3) \\), and \\( l^{prime}(8,-2) \\). complete the following. (a) suppose the angles below have the same measure. choose the correct angle measure. use the protractor provided, as necessary. \\( m \angle j c j^{prime}=m \angle k c k^{prime}=m \angle l c l^{prime}= \\) (b) suppose each pair of segments below have the same length. find each length. give exact answers (not decimal approximations). \\( c j=c j^{prime}= \\) units \\( c k=c k^{prime}= \\) units \\( c l=c l^{prime}= \\) units

Explanation:

Step1: Find the angle measure

When a figure is rotated \(180^{\circ}\) counter - clockwise about the origin, the angle between a point and its image with respect to the origin (the center of rotation) is \(180^{\circ}\). So, \(m\angle JCJ'=m\angle KCK' = m\angle LCL'=180^{\circ}\)

Step2: Use the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For point \(J(-3,7)\) and \(C(0,0)\):

$$
LATEXBLOCK0
$$

Since \(CJ = CJ'\) (rotation preserves distance), \(CJ'=\sqrt{58}\)

For point \(K(-5,3)\) and \(C(0,0)\):

$$
LATEXBLOCK1
$$

Since \(CK = CK'\) (rotation preserves distance), \(CK'=\sqrt{34}\)

For point \(L(-8,2)\) and \(C(0,0)\):

$$
LATEXBLOCK2
$$

Since \(CL = CL'\) (rotation preserves distance), \(CL'=2\sqrt{17}\)

Answer:

(a) \(180^{\circ}\)

(b) \(CJ = CJ'=\sqrt{58}\) units, \(CK = CK'=\sqrt{34}\) units, \(CL = CL'=2\sqrt{17}\) units