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(A) Let $Q(h, k)$ be the image of the point $P(1, 2)$ in the line $x - 3y + 4 = 0$ ... $(1)$.The line $(1)$ is the perpendicular bisector of the line

Vedclass · Accepted Answer

$\left(\frac{6}{5}, \frac{7}{5}\right)$

Vedclass · Answer

(A) Let $Q(h, k)$ be the image of the point $P(1, 2)$ in the line $x - 3y + 4 = 0$ ... $(1)$.The line $(1)$ is the perpendicular bisector of the line segment $PQ$.The slope of the line $x - 3y + 4 = 0$ is $m_1 = \frac{1}{3}$.The slope of the line $PQ$ is $m_2 = \frac{k - 2}{h - 1}$.Since $PQ$ is perpendicular to the line $(1)$,$m_1 	imes m_2 = -1$,so $\frac{1}{3} 	imes \frac{k - 2}{h - 1} = -1$,which gives $k - 2 = -3(h - 1)$,or $3h + k = 5$ ... $(2)$.The midpoint of $PQ$ is $M = \left(\frac{h + 1}{2}, \frac{k + 2}{2}ight)$.Since $M$ lies on the line $(1)$,$\frac{h + 1}{2} - 3\left(\frac{k + 2}{2}ight) + 4 = 0$,which simplifies to $h + 1 - 3k - 6 + 8 = 0$,or $h - 3k = -3$ ... $(3)$.Solving equations $(2)$ and $(3)$:Multiply $(2)$ by $3$: $9h + 3k = 15$.Adding to $(3)$: $(9h + 3k) + (h - 3k) = 15 - 3$,so $10h = 12$,$h = \frac{6}{5}$.Substituting $h$ into $(3)$: $\frac{6}{5} - 3k = -3$,so $3k = \frac{6}{5} + 3 = \frac{21}{5}$,$k = \frac{7}{5}$.Thus,the image is $\left(\frac{6}{5}, \frac{7}{5}ight)$.

Assuming that straight lines work as the plane mirror for a point,find the image of the point $(1, 2)$ in the line $x - 3y + 4 = 0$.

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