The number of integral values of alpha for wh | vedclass

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(B) Given lines are $y = x + 9\alpha$ and $3\alpha x + 2y + 9 = 0$.Substitute $y$ from the first equation into the second:$3\alpha x + 2(x + 9\alpha)

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$1$

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(B) Given lines are $y = x + 9\alpha$ and $3\alpha x + 2y + 9 = 0$.Substitute $y$ from the first equation into the second:$3\alpha x + 2(x + 9\alpha) + 9 = 0$$3\alpha x + 2x + 18\alpha + 9 = 0$$x(3\alpha + 2) = -18\alpha - 9$$x = \frac{-18\alpha - 9}{3\alpha + 2}$To simplify,rewrite the expression for $x$:$x = \frac{-6(3\alpha + 2) + 12 - 9}{3\alpha + 2} = -6 + \frac{3}{3\alpha + 2}$For $x$ to be an integer,$(3\alpha + 2)$ must be a divisor of $3$.The divisors of $3$ are $\pm 1, \pm 3$.Case $1$: $3\alpha + 2 = 1$ $\Rightarrow 3\alpha = -1$ $\Rightarrow \alpha = -1/3$ (not an integer).Case $2$: $3\alpha + 2 = -1$ $\Rightarrow 3\alpha = -3$ $\Rightarrow \alpha = -1$ (integer).Case $3$: $3\alpha + 2 = 3$ $\Rightarrow 3\alpha = 1$ $\Rightarrow \alpha = 1/3$ (not an integer).Case $4$: $3\alpha + 2 = -3$ $\Rightarrow 3\alpha = -5$ $\Rightarrow \alpha = -5/3$ (not an integer).Thus,there is only $1$ integral value of $\alpha$ (which is $\alpha = -1$).

The number of integral values of $\alpha$ for which the abscissa of the point of intersection of the lines $y = x + 9\alpha$ and $3\alpha x + 2y + 9 = 0$ is an integer,is

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