Let alpha in R. If the line (alpha+1) x+alpha | vedclass

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(B) The given equation of the line is $(\alpha+1)x + \alpha y + \alpha - 1 = 0$. Rearranging the terms to group the coefficients of $\alpha$,we get: $

Vedclass · Accepted Answer

$5$

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(B) The given equation of the line is $(\alpha+1)x + \alpha y + \alpha - 1 = 0$. Rearranging the terms to group the coefficients of $\alpha$,we get: $\alpha(x + y + 1) + (x - 1) = 0$. For this line to pass through a fixed point $(h, k)$ for all $\alpha$,the coefficients of $\alpha$ and the constant term must be zero independently: $x + y + 1 = 0$ and $x - 1 = 0$. From $x - 1 = 0$,we get $x = h = 1$. Substituting $x = 1$ into $x + y + 1 = 0$,we get $1 + y + 1 = 0$,which implies $y = k = -2$. Thus,the fixed point is $(1, -2)$. Finally,$h^2 + k^2 = (1)^2 + (-2)^2 = 1 + 4 = 5$.

Let $\alpha \in R$. If the line $(\alpha+1) x+\alpha y+\alpha=1$ passes through a fixed point $(h, k)$ for all $\alpha$,then $h^2+k^2=$

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