Question

worked example
a translation can be measured as a directed line segment.
△mnp was translated to produce △mnp. the triangle was translated a distance equal to the distance between points a and b. it was translated in the direction from point a to point b.
so, $overline{ab}$ is the directed line segment used to measure this translation.

1. suppose each grid square is 1 unit × 1 unit.

a. what is the distance from point a to point b?
b. compare the distance ab with the distances mm, nn, and pp. what do you notice?
c. can you draw another directed line segment on the grid which defines the translation of △mnp to △mnp? if so, draw the segment on the grid. explain your thinking.

1. write equality and congruence statements to compare the corresponding sides and angles of the pre - image △mnp and the image △mnp.

remember:
if △abc≅△def, then:
$overline{ab}congoverline{de}$
$ab = de$
$∠abccong∠def$
$m∠abc=m∠def$

Explanation:

Step1: Use distance - formula for part a

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Count the horizontal and vertical displacements on the grid. Suppose point $A=(x_1,y_1)$ and point $B=(x_2,y_2)$. By counting the grid - squares, if the horizontal displacement $\Delta x=x_2 - x_1 = 4$ units and the vertical displacement $\Delta y=y_2 - y_1=3$ units. Then $d=\sqrt{4^{2}+3^{2}}=\sqrt{16 + 9}=\sqrt{25}=5$ units.

Step2: Analyze translation properties for part b

In a translation, every point of the pre - image is moved the same distance and in the same direction. So, the distances $MM'$, $NN'$, and $PP'$ are all equal to the distance of the translation, which is the distance between $A$ and $B$. So, $AB = MM'=NN'=PP'$.

Step3: Understand translation concept for part c

Yes, we can draw other directed line segments. Any line segment that has the same length and direction as $\overline{AB}$ will define the translation. For example, we can draw a directed line segment from any point on $\triangle MNP$ to its corresponding point on $\triangle M'N'P'$, like from $M$ to $M'$, from $N$ to $N'$, or from $P$ to $P'$. The reason is that in a translation, the vector that describes the movement of one point describes the movement of all points in the figure.

Step4: Write congruence statements for part 3

For the corresponding sides: $\overline{MN}\cong\overline{M'N'}$, $MN = M'N'$; $\overline{NP}\cong\overline{N'P'}$, $NP = N'P'$; $\overline{MP}\cong\overline{M'P'}$, $MP = M'P'$.
For the corresponding angles: $\angle M\cong\angle M'$, $m\angle M=m\angle M'$; $\angle N\cong\angle N'$, $m\angle N=m\angle N'$; $\angle P\cong\angle P'$, $m\angle P=m\angle P'$.

Answer:

a. 5 units
b. $AB = MM'=NN'=PP'$. In a translation, all points of the pre - image are moved the same distance.
c. Yes, we can draw directed line segments from $M$ to $M'$, $N$ to $N'$, or $P$ to $P'$. They have the same length and direction as $\overline{AB}$ which defines the translation.

1. Corresponding sides: $\overline{MN}\cong\overline{M'N'}$, $MN = M'N'$; $\overline{NP}\cong\overline{N'P'}$, $NP = N'P'$; $\overline{MP}\cong\overline{M'P'}$, $MP = M'P'$. Corresponding angles: $\angle M\cong\angle M'$, $m\angle M=m\angle M'$; $\angle N\cong\angle N'$, $m\angle N=m\angle N'$; $\angle P\cong\angle P'$, $m\angle P=m\angle P'$.