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In figure b the point A! is the image of point A under reflection in the mirror m. It is important to realize that the perpendicular distance of point A from the mirror is equal to the perpendicular distance of point A! from the mirror, and the straight line joining A to A! is at right angles to the mirror.
Notation Under a reflection we write that A! is the image of A. If the reflection is called M, then we write M(A) = A!. Alternatively we write: A maps to A! under the reflection M, or in symbols, A —> A! under M.
Sometimes we reflect a shape that lies across the line of the mirror. In figure c, the shaded rectangle has been reflected in the mirror line m.
We can reflect shapes in sloping mirror lines. In figure d, a yacht is reflected in a sloping mirror line m. If we consider the object shape, the image shape, and the mirror line as one composite figure, then the mirror line is an axis of symmetry for this figure.
Example 1. Find the image of the triangle T in figure e after it has been reflected in the two mirrors that are drawn with arrows.
Solution. Draw lines from each vertex of the triangle at right angles to the mirror and produce the lines the same distance on the other side of the mirror. The final image is T". A clockwise rotation about the point where the two mirrors intersect through an angle that is twice the angle between the mirrors will map T onto T”.
See the entry; Composite Transformations for further reflections in two mirrors.
Example 2. Harry lives in a house that is 20 meters from a river, and his uncle’s house is 45 meters from the river (see figure f). These are the shortest distances from the river. The distance apart of the houses, measured along the river bank is 60 meters. Once each week Harry goes down to the river, fills a bucket with river water, and takes it to his uncle’s house to replenish the water in his fish tank. Find the shortest distance that Harry has to walk to accomplish this task.
Solution. Using the bank of the river as a mirror, draw the image of Harry’s house, which is the point B. Join point B to point U with a straight line. The shortest route is from H to P, and then from P to U. This is the same distance as a direct line from B to U. The solution to the problem will be the length of the straight line BU in figure f.
\ / 45 m
20 m /
n \/ C
20 m / 20 m
R 60 m r D
We now extend the line UC (see figure g), which is from uncle’s house to the river, to meet the line through point B and parallel to the bank of the river. In this way the triangle UDB is created, with the length of UD = 65 meters and BD = 60 meters. The route for Harry to follow is from H to P, then from P to U. The length of this route is the same as the length of the straight line BPU, since HP = BP We apply the theorem of Pythagoras to triangle UBD:
UB2 = 652 + 602
UB2 = 4225 + 3600 Squaring 65 and 60
UB2 = 7825
UB = 88.5 (to 1 dp) Using the square root key on the calculator
The shortest distance from Harry’s house to his uncle’s house, via the river, is
Properties of reflection
♦ The mirror line is an invariant line.
♦ Any point on the mirror line is an invariant point.
♦ The perpendicular distance of a point from the mirror line is equal to the perpendicular distance of its image from the mirror line.
♦ The straight line that j oins a point to its image is at right angles to the mirror line.
♦ Angle sizes are invariant.
♦ Area is invariant.
♦ The object and image are congruent shapes.
♦ Reflection is an indirect transformation.
References: Cross-multiply, Composite Transformations, Enlargement, Escher, Image, Indirect Transformation, Perpendicular Lines, Pythagoras’ Theorem, Similar Figures, Symmetry, Transformation Geometry.
Reference: Acute Angle.
References: Euler’s Formula, Networks.
References: Equiangular, Polygon.
Reference: Platonic Solids.
References: Cubic Equations, Quadratic Graphs.
Reference: Recurring Decimal.
Reference: Central Tendency.
RESULTANT OF A VECTOR
Reference: Components of a Vector.
A revolution is a complete turn, which is an angle of 360° or four right angles.
Example 1. If the wheel of a cart has a diameter of 20 cm, find how far the cart travels if the wheel makes one revolution.
Solution. As the wheel makes one revolution, the cart will move forward a distance equal to the circumference of the wheel (see figure). Write
C = 7T x d Formula for circumference of a circle
C = it x 20 Substituting d — 20
C = 62.8 (to 1 dp) Using a scientific calculator
The cart travels a distance of 62.8 cm.