Question

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

Answer 1

The line x+3y=7 acts as a plane mirror.

Let the image of the point A ( 3,8 ) be B ( h,k ).

The line segment passing through the points AB is the perpendicular bisector of the line x+3y=7.

The formula for the slope of a line passing through points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by,

m= y 2 − y 1 x 2 − x 1 (1)

Substitute the values of ( x 1 , y 1 ) and ( x 2 , y 2 ) as ( 3,8 ) and ( h,k ) respectively in equation (1).

m 1 = k−8 h−3

The product of the slopes of two perpendicular lines is given by,

m 1 ⋅ m 2 =−1.(2)

Let m 2 be the slope of line x+3y=7

Rewrite the equation of line x+3y=7.

3y=−x+7 y=− 1 3 x+ 7 3

Compare the above equation with the equation of line y=mx+c.

m 2 =− 1 3

Substitute the value of m 1 , m 2 in equation (2).

k−8 h−3 ×( − 1 3 )=−1 k−8 3h−9 =1 k−8=3h−9 3h−k=1 (3)

Let ( x m , y m ) be the mid-point of the line segment joining points ( 3,8 ) and ( h,k ).

The formula for the mid-point ( x m , y m ) between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by,

( x m , y m )=( x 1 + x 2 2 , y 1 + y 2 2 )(4)

Substitute the values of ( x 1 , y 1 ) and ( x 2 , y 2 ) as ( 3,8 ) and ( h,k ) in equation (4).

( x m , y m )=( h+3 2 , k+8 2 )

The mid-point also satisfies the equation of line x+3y=7.

( h+3 2 )+3( k+8 2 )=7 h+3+3k+24=14 h+3k=14−27 h+3k=−13 (5)

Solve equation (3) and equation (5) to obtain the coordinates of ( h,k ).

Substitute the value of k=3h−1 from equation (3) in equation (5).

h+3( 3h−1 )=−13 h+9h−3=−13 10h=−10 h=−1

Substitute the value of h in equation (3) to obtain the value of k.

k=3×( −1 )−1 =−4

The coordinates of point are ( −1,−4 ).

Thus, the image of point ( 3,8 ) with the line x+3y=7 that acts as a mirror is ( −1,−4 ).