The image of the point (2, 4) with respect to | vedclass

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(A) Let the image of the point $A(2, 4)$ in the line mirror $DE$ be $C(\alpha, \beta)$. Then, $AC$ is perpendicular to $DE$.The midpoint $B$ of $AC$ i

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$\left(-\frac{14}{13}, -\frac{8}{13}\right)$

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(A) Let the image of the point $A(2, 4)$ in the line mirror $DE$ be $C(\alpha, \beta)$. Then, $AC$ is perpendicular to $DE$.The midpoint $B$ of $AC$ is $\left(\frac{\alpha + 2}{2}, \frac{\beta + 4}{2}\right)$.Since point $B$ lies on the line $2x + 3y - 6 = 0$, we have:$2\left(\frac{\alpha + 2}{2}\right) + 3\left(\frac{\beta + 4}{2}\right) - 6 = 0$$\Rightarrow 2\alpha + 4 + 3\beta + 12 - 12 = 0$$\Rightarrow 2\alpha + 3\beta + 4 = 0$  --- $(i)$Since $AC \perp DE$, the product of their slopes is $-1$. The slope of $DE$ is $-\frac{2}{3}$, so the slope of $AC$ must be $\frac{3}{2}$.$\frac{\beta - 4}{\alpha - 2} = \frac{3}{2}$$\Rightarrow 2\beta - 8 = 3\alpha - 6$$\Rightarrow 3\alpha - 2\beta + 2 = 0$  --- $(ii)$Solving equations $(i)$ and $(ii)$:Multiply $(i)$ by $2$ and $(ii)$ by $3$:$4\alpha + 6\beta + 8 = 0$$9\alpha - 6\beta + 6 = 0$Adding these, $13\alpha + 14 = 0 \Rightarrow \alpha = -\frac{14}{13}$.Substituting $\alpha$ in $(i)$:$2(-\frac{14}{13}) + 3\beta + 4 = 0$$-\frac{28}{13} + 3\beta + \frac{52}{13} = 0$$3\beta = -\frac{24}{13} \Rightarrow \beta = -\frac{8}{13}$.Thus, the image of the point $(2, 4)$ is $\left(-\frac{14}{13}, -\frac{8}{13}\right)$.

The image of the point $(2, 4)$ with respect to the straight line $2x + 3y - 6 = 0$ is

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