Question 3

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

Given : $$(8,2)$$ is the solution of equation : $$x+4y-2k=0$$

=> $$8+4(2)-2k=0$$

=> $$2k=8+8=16$$

=> $$k=\frac{16}{2}=8$$

$$\therefore$$ $$k^2=(8)^2=64$$

=> Ans - (C)

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