Find the image of the point (3,8) with respec | vedclass

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(A) The equation of the given line is $x+3y=7$ ... $(1)$.Let point $B(a,b)$ be the image of point $A(3,8)$.Accordingly,line $(1)$ is the perpendicular

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$(-1,-4)$

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(A) The equation of the given line is $x+3y=7$ ... $(1)$.Let point $B(a,b)$ be the image of point $A(3,8)$.Accordingly,line $(1)$ is the perpendicular bisector of $AB$.The slope of $AB = \frac{b-8}{a-3}$,while the slope of line $(1) = -\frac{1}{3}$.Since line $(1)$ is perpendicular to $AB$,we have $\left(\frac{b-8}{a-3}ight) 	imes \left(-\frac{1}{3}ight) = -1$.This simplifies to $\frac{b-8}{3a-9} = 1$,which gives $b-8 = 3a-9$,or $3a-b = 1$ ... $(2)$.The mid-point of $AB$ is $\left(\frac{a+3}{2}, \frac{b+8}{2}ight)$.Since this mid-point lies on line $(1)$,we have $\left(\frac{a+3}{2}ight) + 3\left(\frac{b+8}{2}ight) = 7$.Multiplying by $2$,we get $a+3 + 3b+24 = 14$,which simplifies to $a+3b = -13$ ... $(3)$.Solving equations $(2)$ and $(3)$ simultaneously: From $(2)$,$b = 3a-1$. Substituting into $(3)$,$a + 3(3a-1) = -13$,so $a + 9a - 3 = -13$,$10a = -10$,$a = -1$.Then $b = 3(-1)-1 = -4$.Thus,the image of the point is $(-1,-4)$.

Find the image of the point $(3,8)$ with respect to the line $x+3y=7$,assuming the line to be a plane mirror.

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