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Question

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


Solution


Let line AB be $$x+3y=7$$ and point P be $$(3,8)$$.

Let $$Q(h,k)$$ be the image of point $$P(3,8)$$ in the line $$x+3y=7$$.

Since line $$AB$$ is a mirror,

1) Point $$P$$ and $$Q$$ are at equal distance from line $$AB$$, i.e., $$PR = QR$$, i.e., $$R$$ is the mid-point of $$PQ.$$

2) Image is formed perpendicular to mirror i.e., line PQ is perpendicular to line AB.

Since R is the midpoint of PQ.

Mid point of $$PQ$$ joining $$(3,8)$$ and $$(h,k)$$ is $$\left (\dfrac{h+3}{2},\dfrac{k+8}{2}\right )$$

Coordinate of point R = $$\left (\dfrac{h+3}{2},\dfrac{k+8}{2}\right )$$

Since point R lies on the line AB.
Therefore,
$$\left (\dfrac{3+h}{2}\right )+3\left (\dfrac{8+k}{2}\right )=7$$

$$h+3k=-13$$         ....(1)

Also, $$PQ$$ is perpendicular to $$AB.$$

Therefore,
Slope of $$PQ$$ $$\times$$ Slope of $$AB = -1$$

Since, slope of $$AB =$$ $$-\dfrac{1}{3}$$

Therefore, slope of $$PQ =$$ $$3$$

Now, PQ is line joining $$P(3,8)$$ and $$Q(h,k).$$

Slope of PQ = $$3=\dfrac{k-8}{h-3}$$

$$3h-k=1$$       ........(2)

Solving equation 1 and 2, we get,

$$h=-1$$ and $$k=-4$$

Hence, image is $$Q(-1,-4)$$.

1327170_1052764_ans_4931dea31bff4fd5a1c5b9ee02f7a095.png

Mathematics

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