PLEASE HELPP also sorry the photo is sideways Reflecting /\ (triangle) LMN across the horizontal line y = -1, we get its image /\ (triangle) L' M' N'. Suppose LL', MM', NN' intersect the line of reflection at S, T, and U as shown below.

Accepted Solution

a) Select all that apply[tex]\overline{LL'}, \ \overline{MM'} \ and \ \overline{NN'}[/tex] are each perpendicular to the line of reflectionThis option is the only one that is correct. The line of reflection is  [tex]y=-1[/tex]. When we talk about reflection, we are talking about reflecting across a line, or axis. Reflecting a shape means looking at the mirror image on the other side of the axis. So in this case, this mirror is the line of reflection.  As you can see, these three segments [tex]\overline{LL'}, \ \overline{MM'} \ and \ \overline{NN'}[/tex] form a right angle at the point each segment intersects the line [tex]y=-1[/tex]. b) Find each lengthSince the line [tex]y=-1[/tex] is an axis that allows to get a mirror image, therefore it is true that:[tex]\overline{LS}=\overline{L'S} \\ \\ \overline{MT}=\overline{M'T} \\ \\ \overline{NU}=\overline{N'U}[/tex]To find those values [tex]\overline{LS}[/tex], count the number of units you get from the point S to L, which is 3 units. Do the same to find [tex]\overline{MT}[/tex] but from the point T to M, which is 6 units and finally, for [tex]\overline{NU}[/tex] but from the point U to N, which is 4 units. Therefore:[tex]\overline{LS}=\overline{L'S}=3 \ units \\ \\ \overline{MT}=\overline{M'T}=6 \ units \\ \\ \overline{NU}=\overline{N'U}=4 \ units[/tex]c) Correct StatementThe line of reflection is the perpendicular bisector of each segment joining a point and its image. A bisector is the line dividing something into two equal parts. In this case, the line of reflection divides each segment into two equal parts and is perpendicular because this line form a right angle with each segment. As we demonstrated in a) each segment is perpendicular to the line of reflection, so the first statement is false. On the other hand, each side of the original triangle is not perpendicular to its image and this is obvious when taking a look at the figure. Finally, as we said the line of reflection is perpendicular to each of the mentioned segments, so they can't be parallel as established in the last statement.