Question

Find the image of the point (3, 4) with respect to the line x+3y=7 assuming the line to be a plane mirror​

Answer 1

Answer:

ANSWER

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 (

2

h+3

,

2

k+8

)

Coordinate of point R = (

2

h+3

,

2

k+8

)

Since point R lies on the line AB.

Therefore,

(

2

3+h

)+3(

2

8+k

)=7

h+3k=−13 ....(1)

Also, PQ is perpendicular to AB.

Therefore,

Slope of PQ × Slope of AB=−1

Since, slope of AB= −

3

1

Therefore, slope of PQ= 3

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

Slope of PQ = 3=

h−3

k−8

3h−k=1 ........(2)

Solving equation 1 and 2, we get,

h=−1 and k=−4

Hence, image is Q(−1,−4).