What is the reflection of the point (1, 2) in the line y = 3 ?
(1,-4)
(1,4)
(-1.4)
(-1, 4)
Answers
Answer:
up to top up iiii hjii I am not madan his own way to get to the now
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)