The line (2 + k)x + (1 + k)y = 5 + 7k passes | vedclass

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(D) The given equation is $(2 + k)x + (1 + k)y = 5 + 7k$.Rearrange the equation to group terms with $k$:$2x + kx + y + ky = 5 + 7k$$(2x + y - 5) + k(x

Vedclass · Accepted Answer

$85$

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(D) The given equation is $(2 + k)x + (1 + k)y = 5 + 7k$.Rearrange the equation to group terms with $k$:$2x + kx + y + ky = 5 + 7k$$(2x + y - 5) + k(x + y - 7) = 0$For this line to pass through a fixed point for all values of $k$, both expressions in the parentheses must be zero:$2x + y - 5 = 0$  ---$(1)$$x + y - 7 = 0$  ---$(2)$Subtracting $(2)$ from $(1)$:$(2x + y - 5) - (x + y - 7) = 0$$x + 2 = 0 \implies x = -2$Substitute $x = -2$ into $(2)$:$-2 + y - 7 = 0 \implies y = 9$The fixed point is $(-2, 9)$.The distance $d$ from the origin $(0, 0)$ is $\sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85}$.Therefore, $d^2 = 85$.

The line $(2 + k)x + (1 + k)y = 5 + 7k$ passes through a fixed point for all values of $k$. If $d$ is the distance of this fixed point from the origin, then $d^2 = \dots$

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