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FinnZ [79.3K]
3 years ago
5

The coordinates of the vertices of △JKL△JKL are J(1,4)J(1,4), K(6,4)K(6,4), and L(1,1)L(1,1). The coordinates of the vertices of

△J′K′L′△J′K′L′ are J′(0,−4)J′(0,−4), K′(−5,−4)K′(−5,−4), and L′(0,−1)L′(0,−1). What is the sequence of transformations that maps △JKL△JKL to △J′K′L′△J′K′L′ ?

Mathematics
2 answers:
abruzzese [7]3 years ago
7 0
The sequence of transformations are : reflection across the y axis followed by a reflection across the x axis followed by a translation of 2 units up and 1 unit right.
lianna [129]3 years ago
3 0

Solution: The ΔJKL reflect along the x-axis, then reflects along the y-axis after that translate 1 unit right to transform ΔJ'K'L'.

Explanation:

The vertices of ΔJKL lies in first quadrant because the coordinated of the vertices are positive. It means the vertices are in the form of (x,y).

If the ΔJKL reflects along the x-axis then the vertices are in the form of (x,-y). Now the  ΔJKL reflects along the y-axis then the vertices are in the form of (-x,-y).

The given coordinates of the the vertices after transformation is J'(0,-4), K'(-5,-4), L'(0,-1). But after reflection along x and y axis the vertices of ΔJKL are J'(-1,-4), K'(-6,-4), L'(-1,-1). Since the y coordinates are same so we have to tralanate horizontally. Since the difference between x coordinates is positive 1, so we have translate the triangle 1 unit right.

Therefore, the ΔJKL reflect along the x-axis, then reflects along the y-axis after that translate 1 unit right to transform ΔJ'K'L'.

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Step-by-step explanation:

Given

Let:

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Solve for A, C and S

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A = \frac{C}{2} -- (3)

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Make S the subject

S = 329 - 3A

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S = 329 - 3 * 85

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Hence, the result is:

C = 170

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A = 85

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