Answer:
[tex]Solution,\\PQ = \sqrt{(1-3)^2+(3-1)^2} = \sqrt{2^2+2^2} = \sqrt{8} = 2\sqrt{2} \\P'Q' = \sqrt{(3-9)^2+(9-3)^2} = \sqrt{6^2+6^2} = \sqrt{72} = 6\sqrt{2}\\Also,\\RQ = \sqrt{(1-3)^2+(1-1)^2} = \sqrt{2^2} = 2\\R'Q' = \sqrt{(3-9)^2+(3-3)^2} = \sqrt{6^2} =6\\Again,\\PR = \sqrt{(1-1)^2+(3-1)^2} = \sqrt{2^2}=2\\P'R' = \sqrt{(3-3)^2+(9-3)^2} = \sqrt{6^2} = 6\\Here, P'R' = 3PR, P'Q'=3PQ ~and~ Q'R'=3RQ~\\In~other~words~each~side~of~triangle~P'Q'R' ~is~three~times~the~length~of~triangle~PQR.\\[/tex][tex]So,~the~algebraic~rule~of~diliation(enlargement)~is~(x,y)----(3x,3y)[/tex]
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