1. If (8, 2) is a solution of x+4y-2k=0 then find the value of $$k^2$$.





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  • By: anil on 05 May 2019 02.03 am
    Given : $$(8,2)$$ is the solution of equation : $$x+4y-2k=0$$ => $$8+4(2)-2k=0$$ => $$2k=8+8=16$$ => $$k=frac{16}{2}=8$$ $$ herefore$$ $$k^2=(8)^2=64$$ => Ans - (C)
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