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GuDViN [60]
3 years ago
9

Need help in geometry!!!

Mathematics
1 answer:
Ludmilka [50]3 years ago
7 0
Given:
JKLM: J(8,4) ; K(4,10) ; L(12,12) ; M(14,10) 

I'll just use the first vertex of every quadrilateral in comparison with J(8,4)

A(-8,-4) ; B(-4,-10) ; C(-12,-12) ; D(-14,-10) → a sequence of reflections across the x and y-axes, in any order. This is easy because each vertex has the same number but different signs. 

E(11,2) ; F(7,8) ; G(15,10) ; H(17,8) → a translation of 3 units left and 2 units up. As you can envision with a graph, from 11 to 8, you need to move to the left of the x-axis. from 2 to 4, you need to move up of the y-axis.

O(6,7) ; P(2,13) ; Q(10,15) ; R(12,13) → a translation 2 units right and 3 units down. From 6 to 8 of the x-axis and from 7 to 4 of the y axis. 

S(4,16) ; T(10,20) ; U(12,12) ; V(10,10) → This one has no pair but it seems as though this is rotated along the point U.

W(11,7) ; X(7,13) ; Y(15,15) ; Z(17,13) → a translation 3 units down and 3 units left. 
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marin [14]

Answer:

        Co ordinate of A' = (1, -4)

        Co ordinate of B' = (3, 2)

        Co ordinate of C' = (4, -2)

Explanation:

 The figure of the triangle is shown below.

  If we are finding image of the triangle about X-axis all the x -co-ordinates will remain same, but y co-ordinate will change it's sign.

 So, co ordinate of A' = (1, -4)

        Co ordinate of B' = (3, 2)

        Co ordinate of C' = (4, -2)

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