When P(x, y) is reflected in the mirror line to become P(x', y') the mirror line perpendicularly bisects [pp]'
Thus every point on an object, the mirror line perpendicularly bisects the line segment joining the point with its image.
Reflection in the x-axis Reflection in the y-axis Reflection in the line y = x Reflection in the line y = -x |
(x, y) --> (x, -y) (x, y) --> (-x, y) (x, y) --> (y, x) (x, y) --> (-y, -x) |
Problem 1 :
Find the image equation of 2x - 3y = 8 reflected in the y-axis.
Solution :
Reflection in the y-axis :
Rule to be applied :
x should be changed as -x, y will be the same.
2(-x) - 3y = 8
-2x - 3y = 8
Multiplying by negative, we get
2x + 3y = -8
Problem 2 :
Find the image equation of y = 2x +3 under M_{x}
Solution :
Reflection under the x-axis :
Rule to be applied :
y should be changed as -y, x will be the same.
-y = 2x +3
2x + y + 3 = 0
or
y = -2x - 3
Problem 3 :
Find the image equation of y = x^{2} under M_{x}
Solution :
Reflection under the x-axis :
Rule to be applied :
y should be changed as -y, x will be the same.
-y = x^{2}
y = -x^{2}
Problem 4 :
Find the image equation of y = 2^{x} under M_{y = x}
Solution :
Reflection under the x-axis :
Rule to be applied :
Change x as y and change y as x.
x = 2^{y}
Problem 5 :
Find the image equation of 2x + 3y = 4 under M_{y = -x}
Solution :
Reflection under the line y = -x :
Rule to be applied :
Change x as -y and change y as -x.
2(-y) + 3(-x) = 4
-2y - 3x = 4
Multiplying by negative sign, we get
3x + 2y = -4
Problem 6 :
Find the image equation of x = 3 under M_{y = -x}
Solution :
Reflection under the line y = -x :
Rule to be applied :
Change x as -y and change y as -x.
-y = 3
y = -3
Problem 7 :
Find the image equation of y = 2x^{2} under M_{y = x}
Solution :
Reflection under the line y = x :
Rule to be applied :
Change x as y and change y as x.
x = 2y^{2}
Problem 8 :
Find the image equation of y = 5/x under M_{y = x}
Solution :
Reflection under the line y = x :
Rule to be applied :
Change x as y and change y as x.
x = 5/y
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