Question

The reflection of point A(4,-1)in the line x=2

Answer 1

Answer:

1,2

3,4

−1,0-1,0

2,3

Answer :

D

Solution :

Let Q(n,k) be the image of the point P (4,1) to the line k -n +1 =0 . Then PQ is perpendicular to k -n +1 =0

k−1n−4×1=−1k-1n-4×1=-1

k +n =4 +1=5

Also , mid-point of PQ i.e,(4+n2,k+12)(4+n2,k+12) lies on k-n+1=0

k−1n−4×1=−1k-1n-4×1=-1

k +n=4+1 =5

Also, mid-point of PQ , i.e(4+n2,k+12)(4+n2,k+12) lies on

k-n +1 =0

k+12,4+n2+1=0k+12,4+n2+1=0

k -n-1=0

on solving Eqs. (i) and (ii) we get the required point (2,3)

Answer 2

(0,-1) is the reflection of point A(4,-1) in the line x=2.

Given:

Point A(4,-1) and a line x=2.

To Find:

The reflection of point A(4,-1) on the line x=2.

Solution:

The horizontal distance between the line x=2 and point A(4,-1) is 2 units.

The reflection of the point A(4,-1) on the line x=2 will be at the horizontal distance of 2 units to the left of the line x=2. So, that reflected point is (0,-1). The $y$-coordinate is the same here because the is asking about a line parallel to the y-axis.

This can also be understood from the graph. Look at the graph attached below.

Thus, (0,-1) is the reflection of point A(4,-1) in the line x=2.