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 $$Q(-1,-4)$$.Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More