Question

Point A(15, –2) is reflected in x-axis as A’, point B(–3, –3) is reflected in the y-axis as B’. Find the distance between A’ and B’

Answer 1

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Given,

P’ (2, -3) is the reflection of P (a, b) in the x-axis

Hence, the co-ordinates of P’ will be (a, – b) but P’ is (2, -3)

On comparing, we get a = 2, b = 3

Thus, the co-ordinates of P will be (2, 3)

And,

P” is the image of P when reflected in y-axis

Hence, the co-ordinate of P” will be ( – 2, 3)

Now, draw a line x = 4, which is parallel to y-axis

As P’” is the image of P when it is reflected in the line x = 4,

So, P’” is its reflection.

Thus, the co-ordinates of P”’ will be (6, 3).

Answer 2

SOLUTION

GIVEN

Point A(15, –2) is reflected in x-axis as A' , point B(–3, –3) is reflected in the y-axis as B'

TO DETERMINE

The distance between A' and B'

CONCEPT TO BE IMPLEMENTED

Let a given point is (a, b)

• If it reflected in x-axis then the coordinates is (a, - b)

• If it reflected in y-axis then the coordinates is ( - a, b)

EVALUATION

Here it is given that point A(15, –2) is reflected in x-axis as A'

So coordinate of A' is ( 15, 2)

Again point B(–3, –3) is reflected in the y-axis as B'

So coordinate of B' is ( 3 , - 3)

Hence the required distance between A' and B'

= Distance between ( 15, 2) and ( 3 , - 3)

$\sf = \sqrt{ {(15 - 3)}^{2} + {(2 + 3)}^{2} }$

$\sf = \sqrt{ {(12)}^{2} + {(5)}^{2} }$

$\sf = \sqrt{144 + 25}$

$\sf = \sqrt{169}$

$\sf = 13$

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