Question

reflecting at mn across the horizontal line ( y = 1 ), we get its image ( \triangle mn ).
suppose ( ll ), ( mm ), and ( nn ) intersect the line of reflection at ( s ), ( t ), and ( u ) as shown below.
complete the following.
(a) select all that apply.
( ll ) is perpendicular to ( mm ) and ( nn ).
( ll ), ( mm ), and ( nn ) are each perpendicular to the line of reflection.
( ll ), ( mm ), and ( nn ) are parallel to each other.
none of the above
(b) suppose each pair of segments below have the same length.
find each length.
( ls = ls = square ) units
( mt = mt = square ) units
( nu = nu = square ) units
(c) choose the correct statement about the reflection.
the line of reflection is the perpendicular bisector of each segment joining a point and its image.
the line of reflection is parallel to each segment joining a point and its image.
the line of reflection is neither parallel nor perpendicular to each segment joining a point and its image.
each side of the original figure is perpendicular to its image.

Explanation:

(a)

• Step1: Analyze the property of reflection

When a figure is reflected over a line, the line of reflection is the perpendicular bisector of the segment joining a point and its image. For a horizontal line of reflection \(y = 1\), if we consider two - point segments \(LT\), \(MW\), and \(NV\):

• The line of reflection \(y = 1\) is not perpendicular to \(MW\) and \(NV\) (because \(MW\) and \(NV\) are not vertical lines). But \(LT\) is a horizontal segment (since the \(y\) - coordinates of \(L\) and \(T\) are the same). The line of reflection \(y = 1\) is perpendicular to \(LT\) (because a horizontal line \(y = 1\) and a horizontal segment \(LT\) are perpendicular in the sense of the reflection property where the line of reflection is the perpendicular bisector of the segment joining a pre - image and its image).

(b)

• Step1: Use the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
If \(L(-8,2)\) and \(L'(- 2,2)\) (since reflection over \(y = 1\) for a horizontal segment \(LT\) (assuming \(T\) and \(T'\) are related in the same way as \(L\) and \(L'\)):
\(LS=\vert-8-(-2)\vert\div2\). The \(x\) - coordinate of \(L\) is \(x=-8\) and the \(x\) - coordinate of \(L'\) is \(x'=-2\). The distance between \(L\) and \(L'\) along the \(x\) - axis (since \(y\) - coordinates are the same) is \(d=\vert-8 + 2\vert=\vert-6\vert = 6\). Since \(S\) is the mid - point of \(LL'\) (because the line of reflection \(y = 1\) is the perpendicular bisector of \(LL'\)), \(LS=\frac{\vert-8-(-2)\vert}{2}=3\) units.

• Step2: Use the distance formula for \(MT\)

If \(M(-8,0)\) and \(M'(-2,0)\) (using the same reflection principle). The distance between \(M\) and \(M'\) along the \(x\) - axis (since \(y\) - coordinates are \(0\)) is \(d=\vert-8+2\vert = 6\). So \(MT = 3\) units.

• Step3: Use the distance formula for \(NU\)

If \(N(-8,4)\) and \(N'(-2,4)\) (using the reflection property). The distance between \(N\) and \(N'\) along the \(x\) - axis (since \(y\) - coordinates are \(4\)) is \(d=\vert-8 + 2\vert=6\). So \(NU = 3\) units.

(c)

• Step1: Recall the property of reflection

The line of reflection is the perpendicular bisector of each segment joining a point and its image.

Answer:

(a) \(LT\) is perpendicular to \(MW\) and \(NV\).
(b) \(LS = 3\) units, \(MT=3\) units, \(NU = 3\) units.
(c) The line of reflection is the perpendicular bisector of each segment joining a point and its image.