Math, asked by vijaysaradhi2015, 6 months ago

What is the reflection of the point (1, 2) in the line y = 3 ?
(1,-4)
(1,4)
(-1.4)
(-1, 4)​

Answers

Answered by Nandanshinganalli
0

Answer:

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Answered by amitkumarswain2005
0

Answer:

The line y = 3 is parallel to x-axis.

The line y = 3 is parallel to x-axis.Let the required image is P′

The line y = 3 is parallel to x-axis.Let the required image is P′By common sense, we know

The line y = 3 is parallel to x-axis.Let the required image is P′By common sense, we know(Distance between the line y = 3 and point P) = (Distance between line y= 3 and point P′)

The line y = 3 is parallel to x-axis.Let the required image is P′By common sense, we know(Distance between the line y = 3 and point P) = (Distance between line y= 3 and point P′)Since line joining PP′ is perpendicular to y = 3, the x- coordinates of P′ will remain 1. The y coordinate must be 4 because the distance between line and P is equal to 3 - 2 = 1).

The line y = 3 is parallel to x-axis.Let the required image is P′By common sense, we know(Distance between the line y = 3 and point P) = (Distance between line y= 3 and point P′)Since line joining PP′ is perpendicular to y = 3, the x- coordinates of P′ will remain 1. The y coordinate must be 4 because the distance between line and P is equal to 3 - 2 = 1).Reflected point P′ should also be 1 unit away from the line making y-coordinate (3 + 1 = 4).

The line y = 3 is parallel to x-axis.Let the required image is P′By common sense, we know(Distance between the line y = 3 and point P) = (Distance between line y= 3 and point P′)Since line joining PP′ is perpendicular to y = 3, the x- coordinates of P′ will remain 1. The y coordinate must be 4 because the distance between line and P is equal to 3 - 2 = 1).Reflected point P′ should also be 1 unit away from the line making y-coordinate (3 + 1 = 4).Required point = P′ = (1,4)

OR

The line y = 3 bisects the line joining P and P′.

The line y = 3 bisects the line joining P and P′.The midpoint of these two points must lie on the line y = 3.

The line y = 3 bisects the line joining P and P′.The midpoint of these two points must lie on the line y = 3.If we take point P′ as (1,b), the midpoint of P & P’ (1 + 1)/2 , (b + 2)/2 must satisfy the equation y = 3.

The line y = 3 bisects the line joining P and P′.The midpoint of these two points must lie on the line y = 3.If we take point P′ as (1,b), the midpoint of P & P’ (1 + 1)/2 , (b + 2)/2 must satisfy the equation y = 3.=> (b + 2)/2 = 3

The line y = 3 bisects the line joining P and P′.The midpoint of these two points must lie on the line y = 3.If we take point P′ as (1,b), the midpoint of P & P’ (1 + 1)/2 , (b + 2)/2 must satisfy the equation y = 3.=> (b + 2)/2 = 3=> b = 6 - 2 = 4

The line y = 3 bisects the line joining P and P′.The midpoint of these two points must lie on the line y = 3.If we take point P′ as (1,b), the midpoint of P & P’ (1 + 1)/2 , (b + 2)/2 must satisfy the equation y = 3.=> (b + 2)/2 = 3=> b = 6 - 2 = 4Required point is (1,4)

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