Question

1. draw the image of quadrilateral abcd when translated by the directed line segment t. label the image of a as a, the image of b as b, the image of c as c, and the image of d as d. (from unit 1, lesson 12.) 5. here is a line $ell$. plot 2 points, a and b, which stay in the same place when they are reflected over $ell$. plot 2 other points, c and d, which move when they are reflected over $ell$. (from unit 1, lesson 11.) 6. here are 3 points in the plane. select all the straightedge and compass constructions needed to locate the point that is the same distance from all 3 points. a. construct the bisector of angle cab. b. construct the bisector of angle cba. c. construct the perpendicular bisector of bc. d. construct the perpendicular bisector of ab. e. construct a line perpendicular to ab through point c. f. construct a line perpendicular to bc through point a. (from unit 1, lesson 9.)

Explanation:

Step - by - Step Format

1. Question 4:

• Step1: Understand translation
• A translation moves every point of a figure the same distance and in the same direction as the given directed - line segment \(t\). To find the image of point \(A\) (say \(A'\)), measure the length and direction of \(t\) from \(A\) and mark \(A'\) at the end of that displacement. Repeat this process for points \(B\), \(C\), and \(D\) to get \(B'\), \(C'\), and \(D'\) respectively.

1. Question 5:

• Step1: Identify invariant points
• Points \(A\) and \(B\) are invariant under reflection over line \(\ell\), so they remain in the same place.
• Step2: Reflect non - invariant points
• To reflect points \(C\) and \(D\) over line \(\ell\), draw perpendiculars from \(C\) and \(D\) to line \(\ell\). Measure the distance from \(C\) (or \(D\)) to the foot of the perpendicular on \(\ell\). Then, on the other side of \(\ell\), mark a point at the same distance from the foot of the perpendicular to get the reflected points \(C'\) and \(D'\).

1. Question 6:

• Step1: Recall the circum - center property
• The point that is equidistant from three non - collinear points \(A\), \(B\), and \(C\) is the circum - center of the triangle formed by these three points. The circum - center is the intersection of the perpendicular bisectors of the sides of the triangle.
• Step2: Select the correct construction steps
• To find the point equidistant from \(A\), \(B\), and \(C\), we need to construct the perpendicular bisector of \(AB\) (option D) and the perpendicular bisector of \(BC\) (option C). The intersection of these two perpendicular bisectors is the point that is equidistant from \(A\), \(B\), and \(C\).

Answer:

1. Follow the translation steps as described above to draw \(A'\), \(B'\), \(C'\), and \(D'\).
2. Plot \(A\) and \(B\) in the same place, and reflect \(C\) and \(D\) over line \(\ell\) as described above.
3. C. Construct the perpendicular bisector of \(BC\); D. Construct the perpendicular bisector of \(AB\).