Question

Find the image of point P(1,-2) in a line 4x = 3y + 5.

Answer 1

Step-by-step explanation:

Let line AB be 4x= 3y + 5 and point P be (1,-2).

Let Q(h,k) be the image of point P(1,-2) in the line

4x= 3y + 5.

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 (1,-2) and (h,k) is

[ (1 + h ) /2 , (-2 + k)/2 ]

Coordinate of point R = [ (1 + h ) /2 , (-2 + k)/2 ]

Since point R lies on the line AB.

Therefore,

4(1 + h ) /2 = 3 (-2 + k)/2 + 5

4+ 4h = - 6 + 3k + 10

4h - 3k = 0 ..... (1)

Also, PQ is perpendicular to AB.

Therefore,

Slope of PQ × Slope of AB=−1

Since, slope of AB= − 1/3

Therefore, slope of PQ= 3

Now, PQ is line joining P(1,-2) and Q(h,k).

Slope of PQ =

3= h−1/ k + 2

3k + 6 = h - 1

3k - h = -7 ........(2)

Solving equation 1 and 2, we get,

h=−7/3 and k= -28/9

Hence, image is Q(−7/3,−28/9).