The value of a for which the image of the poi | vedclass

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(C) Let the point $P = (a, a - 1)$ and its image $Q = (a^2 + 1, a)$ with respect to the line $3x + y - 6a = 0$.$1$. The midpoint $L$ of $PQ$ must lie

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$2$

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(C) Let the point $P = (a, a - 1)$ and its image $Q = (a^2 + 1, a)$ with respect to the line $3x + y - 6a = 0$.$1$. The midpoint $L$ of $PQ$ must lie on the line $3x + y = 6a$.Midpoint $L = \left( \frac{a + a^2 + 1}{2}, \frac{a - 1 + a}{2} ight) = \left( \frac{a^2 + a + 1}{2}, \frac{2a - 1}{2} ight)$.Substituting $L$ into the line equation: $3\left( \frac{a^2 + a + 1}{2} ight) + \left( \frac{2a - 1}{2} ight) = 6a$.$3a^2 + 3a + 3 + 2a - 1 = 12a \Rightarrow 3a^2 - 7a + 2 = 0$.$(3a - 1)(a - 2) = 0 \Rightarrow a = 2$ or $a = 1/3$.$2$. The line $PQ$ must be perpendicular to the mirror line $3x + y = 6a$.Slope of $PQ = \frac{a - (a - 1)}{a^2 + 1 - a} = \frac{1}{a^2 - a + 1}$.Slope of mirror line is $-3$.Since $PQ \perp 	ext{mirror}$,product of slopes is $-1$: $\left( \frac{1}{a^2 - a + 1} ight) 	imes (-3) = -1$.$a^2 - a + 1 = 3 \Rightarrow a^2 - a - 2 = 0$.$(a - 2)(a + 1) = 0 \Rightarrow a = 2$ or $a = -1$.Comparing both conditions,the common value is $a = 2$.

The value of $a$ for which the image of the point $(a, a - 1)$ with respect to the line mirror $3x + y = 6a$ is the point $(a^2 + 1, a)$ is:

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