Using a graph paper, plot the point A (6, 4) and B (0, 4).

(a) Reflect A and B in the origin to get the image A’ and B’.

(b) Write the co-ordinates of A’ and B’.

(c) Sate the geometrical name for the figure ABA’B’.

(d) Find its perimeter.

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#### Solution

(a)

(b) Co-ordinates of A'=(-6,-4)

Co-ordinates of B'=(0,-4)

(c) ABA'B' is parallelogram.

(d) In ABA'B',BB'=8 units, A'B'=6 units

`therefore BA'=sqrt(6^2+8^2)=sqrt(36+64)=sqrt100=10 `units

`=>B'A=10 ` units

AB=A'B'=6units

∴ the perimeter of ABA'B'=AB+BA'+A'B'+B'A

=6+10+6+10

=32 units

Concept: Invariant Points.

Is there an error in this question or solution?

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