Question

practice
1 write a function to describe each translation. each solid figure is a pre - image, and each dashed figure is an image.
(a) image of a parallelogram translation with points a and b
(b) image of a trapezoid - like figure translation with points a and b
(c) image of a hexagon translation with points p and q
(d) image of a triangle translation with points p and q
© carnegie learning, inc.

Explanation:

Step1: Analyze Translation Vector

For a translation, the function is determined by the vector from a pre - image point to its image point. Let's take part (a) as an example. Let point \(A\) be a point on the pre - image (solid figure) and point \(B\) be the corresponding point on the image (dashed figure). The translation vector \(\vec{v}=\langle x_2 - x_1,y_2 - y_1 angle\), where \((x_1,y_1)\) are the coordinates of \(A\) and \((x_2,y_2)\) are the coordinates of \(B\). From the diagram, if we assume a coordinate system, the horizontal (x - direction) and vertical (y - direction) changes can be observed. The translation from \(A\) to \(B\) gives the vector for the translation of the entire figure. The general form of a translation function in 2 - D is \((x,y)\to(x + h,y + k)\), where \((h,k)\) is the translation vector.

Step2: Write the Function for Each Part

Part (a)

Looking at the vector from \(A\) to \(B\), if we assume the horizontal shift is \(h\) and vertical shift is \(k\). From the diagram, the pre - image (solid parallelogram) is translated to the image (dashed parallelogram) in the same direction as the vector from \(A\) to \(B\). If we consider the coordinates, suppose \(A=(x_1,y_1)\) and \(B=(x_1 + h,y_1 + k)\). From the visual, the horizontal shift \(h\) and vertical shift \(k\) can be determined. Let's assume that the translation vector is \(\langle a,b angle\) (by observing the distance and direction from \(A\) to \(B\)). The translation function is \((x,y)\to(x + a,y + b)\). For example, if \(A\) moves right by some units and up by some units, the function will reflect that.

Part (b)

For the trapezoid, the vector from \(B\) (on pre - image) to \(A\) (on image) gives the translation vector. Let the coordinates of \(B=(x_1,y_1)\) and \(A=(x_2,y_2)\). Then \(h=x_2 - x_1\) and \(k=y_2 - y_1\). The translation function is \((x,y)\to(x + h,y + k)\).

Part (c)

For the hexagon, the vector from the point on the pre - image (solid hexagon) to the corresponding point on the image (dashed hexagon) and also the vector from \(Q\) to \(P\) (since the translation of the hexagon is the same as the translation of the segment \(QP\) in reverse? Wait, no. The pre - image is the solid hexagon and the image is the dashed hexagon below? Wait, no, the solid is pre - image, dashed is image. Wait, in part (c), the solid hexagon is above and the dashed is below? Wait, no, the problem says "each solid figure is a pre - image, and each dashed figure is an image". So solid is pre, dashed is image. So the translation is from solid (pre) to dashed (image). The segment \(QP\): \(Q\) is on pre - image related? Wait, the solid hexagon is translated to the dashed hexagon. The vector from a point on solid hexagon to dashed hexagon is the same as the vector from \(Q\) to \(P\) but in reverse? Wait, no, \(P\) and \(Q\): if \(Q\) is on pre - image and \(P\) is on image? Wait, the diagram shows \(Q\) below and \(P\) above. Wait, maybe the translation is vertical. Let's say the solid hexagon is translated down (since the dashed is below). The translation vector is \(\langle0,-m angle\) (where \(m\) is the vertical distance). So the function is \((x,y)\to(x,y - m)\).

Part (d)

For the triangle, the pre - image is the dashed triangle and the image is the solid triangle? Wait, no: "each solid figure is a pre - image, and each dashed figure is an image". Wait, no, the problem says "each solid figure is a pre - image, and each dashed figure is an image". So solid is pre, dashed is image. Wait, in part (d), the solid triangle is pre - image? No, the dashed triangle is image? Wait, the diagram: the dashed triangle is above and to the right, and the solid is below and to the right? Wait, the point \(P\): the pre - image point (on solid) and image point (on dashed). The vector from \(P\) (on solid) to the corresponding point on dashed. The translation vector: horizontal and vertical. If we look at the dashed triangle and solid triangle, the translation is such that the horizontal shift and vertical shift can be determined from the position of \(P\) and its image. The function is \((x,y)\to(x + h,y + k)\) where \(h\) and \(k\) are the horizontal and vertical shifts.

Let's take a more concrete approach. For a general translation, if we have a pre - image point \((x,y)\) and the image point \((x',y')\), then \(x'=x + h\) and \(y'=y + k\), so the translation function is \(T(x,y)=(x + h,y + k)\).

Example for part (a)

Suppose the vector from \(A\) to \(B\) is \(\langle3,2 angle\) (by counting units, assuming a grid). Then the translation function is \((x,y)\to(x + 3,y + 2)\).

Example for part (b)

If the vector from \(B\) to \(A\) is \(\langle - 2,4 angle\) (wait, no, \(B\) is on pre - image, \(A\) on image. So the vector from \(B\) to \(A\) is \((x_A - x_B,y_A - y_B)\). So the translation function is \((x,y)\to(x+(x_A - x_B),y+(y_A - y_B))\).

Part (c)

The segment \(QP\): \(Q\) is below, \(P\) is above. But the pre - image (solid hexagon) is translated to the image (dashed hexagon) which is below the solid one. So the translation vector is \(\langle0,-n angle\) (vertical down). So the function is \((x,y)\to(x,y - n)\).

Part (d)

The solid triangle is pre - image, dashed is image? Wait, no, the problem says solid is pre, dashed is image. Wait, in part (d), the dashed triangle is image, solid is pre. The point \(P\): the pre - image \(P\) and image \(P\) (dashed). The translation vector from pre - image \(P\) to image \(P\) is horizontal and vertical. If the pre - image \(P\) is at \((x,y)\) and image \(P\) is at \((x + h,y + k)\), then the function is \((x,y)\to(x + h,y + k)\).

Answer:

The translation functions are as follows:

Part (a)

Let the translation vector from a pre - image point to image point be \(\langle h,k
angle\). If we observe the vector from \(A\) to \(B\), the translation function is \(\boldsymbol{(x,y)\to(x + h,y + k)}\) (where \(h\) and \(k\) are the horizontal and vertical shifts determined by the vector \(\overrightarrow{AB}\)).

Part (b)

Let the translation vector from a pre - image point (on the trapezoid) to its image point (on the dashed trapezoid) be \(\langle h,k
angle\) (determined by the vector \(\overrightarrow{BA}\)). The translation function is \(\boldsymbol{(x,y)\to(x + h,y + k)}\).

Part (c)

The translation is vertical. Let the vertical shift be \(k\) (negative since it's a downward shift). The translation function is \(\boldsymbol{(x,y)\to(x,y + k)}\) (where \(k\) is negative, e.g., if the shift is 3 units down, \(k=- 3\)).

Part (d)

Let the horizontal shift be \(h\) and vertical shift be \(k\) determined by the position of \(P\) (pre - image) and its image (on the dashed triangle). The translation function is \(\boldsymbol{(x,y)\to(x + h,y + k)}\).

(Note: To get the exact values of \(h\) and \(k\) for each part, we need to use a coordinate system and measure the distance between corresponding points. For example, if we place \(A\) at \((0,0)\) and \(B\) at \((2,1)\) in part (a), then the translation function is \((x,y)\to(x + 2,y + 1)\))