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Damm [24]
3 years ago
12

HELP!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Olegator [25]3 years ago
5 0

Answer:

Part) A. Part B and Part C see the explanation

Step-by-step explanation:

Part A) Translate Triangle ABC 6 units horizontally How are the values in the ordered pairs affected by the translation?

If the translation is 6 units horizontally, that means that each vertex of the ABC pre-image is going to move 6 units to the right, i.e. each x-coordinate of the ABC pre-image is going to increase by 6 units and each y-coordinate of the ABC pre-image is going to remain the same.

Part B) Translate Triangle ABC -3 units vertically How are the values in the ordered pairs affected by the translation?

If the translation is -3 units vertically, that means that each vertex of the ABC pre-image is going to move 3 units down, i.e. each x-coordinate of the ABC pre-image is going to remain the same and each y-coordinate of the ABC pre-image is going to decrease by 6 units.

Part C) How could you determine the coordinates of the vertices of a translated image without using a graph?

we know that

The coordinates of the image vertices can be determined through the rule of the translation

so

In the part A) the translation is +6 units horizontally (6 units at right)

therefore

The rule of the translation of the pre-image to the image is equal to

(x,y) ------> (x+6,y)

<u><em>Example</em></u>

The coordinates of the pre-image A(-4,1)

To obtain the coordinates of the image A' apply the rule of the translation

A(-4,1)--------> A'(-4+6,1)

A(-4,1)--------> A'(2,1)

In the part B) the translation is -3 units vertically (3 units down)

therefore

The rule of the translation of the pre-image to the image is equal to

(x,y) ------> (x,y-3)

<u><em>Example</em></u>

The coordinates of the pre-image A(-4,1)

To obtain the coordinates of the image A' apply the rule of the translation

A(-4,1)--------> A'(-4,1-3)

A(-4,1)--------> A'(-4,-2)

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Since a rhombus is a parallelogram.

By property of rhombus , if point N is the intersection of the diagonals as shown in the figure, then

\overline{MN} \cong \overline{NK}         .....[1]

\overline{JN} \cong \overline{NL}

In ΔJNM and ΔJNK

\overline{MN} \cong \overline{NK}     [side]    [by (1)]

\overline{JM} \cong \overline{JK}     [side]          [Given]

By reflexive property states that a segment is congruent to itself:

\overline{JN} \cong \overline{JN}   [Side]                 [Reflexive Property]

SSS(Side-Side-Side) postulates states that if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent.

then by SSS congruence,

\triangle JNM \cong \triangle JNK

By CPCT [Corresponding Part of congruent triangles are congruent]

Since, JNM and JNK are corresponding angles therefore,

\angle JNM \cong JNK

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By linear pair theorem, JNM and JNK are supplementary

this mean:

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Since, the angles are congruent i.e, \angle JNM \cong JNK

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or

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2m\angle JNK=180^{\circ}

Simplify:

m\angle JNK =90^{\circ}

also; m\angle JNM =90^{\circ}

therefore, the diagonals of JKLM are perpendicular to each other i,e \overline{JL} \perp \overline{MK}


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