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Shape P has been shown on the grid below.
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Part a) Reflect shape P in the đť‘Ą-axis.
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In the first part of this question, weâ€™ve been asked to perform a reflection and the mirror line that weâ€™re using is the đť‘Ą-axis; thatâ€™s this line here.
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When we perform a reflection, the image of the shape that weâ€™re reflecting will end up on the opposite side of the mirror line.
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But all of the corners of the image will be the same distance from the mirror line as they were on the original shape.
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We need to work perpendicular or at right angles to the mirror line.
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Letâ€™s start with this corner here.
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We can see that this corner is one square vertically below the mirror line.
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So the image of this corner will be one square vertically above the mirror line because itâ€™s on the opposite side.
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So this corner will be here on the image.
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This corner is also one square vertically below the mirror line.
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So its image will also be one square vertically above the mirror line.
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This corner is four squares vertically below the mirror line.
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So its image will be four squares vertically above the mirror line.
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Weâ€™ve now found the positions of three of the corners of the shape on the image.
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Weâ€™ve just got two left.
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This corner is two squares vertically below the mirror line.
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So on the image, it will be two squares vertically above the mirror line and, finally, this corner, which is four squares vertically below the mirror line.
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So its image will be four squares vertically above the mirror line.
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To complete the question then, we connect the five corners together with straight lines and we can shade the inside of the shape, if we want.
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We can see then that the image of this shape after itâ€™s been reflected in the đť‘Ą-axis is exactly the same size and shape as the original shape P, but is in a different orientation; itâ€™s upside down relative to the original shape.
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This is consistent with what we see when we look in the mirror: our left hand becomes our reflectionâ€™s right hand and so on.
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The semicircles Q and R have been shown on the grid below.
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Part b) Fully describe the single transformation that maps the semicircle Q onto the semicircle R.
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There are four types of transformation that we need to consider.
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They are translation, rotation, reflection, and enlargement.
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Weâ€™ve been specifically asked for the single transformation that maps Q onto R, which means we need to be able to get there in one step.
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We canâ€™t give a combination of transformations as our answer.
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Letâ€™s consider whether the transformation that weâ€™re looking for could be a translation first of all.
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Now, a translation is just a shift or a movement of a shape.
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The image of the shape is exactly the same size and shape and itâ€™s in the same orientation; itâ€™s just in a different position on the coordinate grid.
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If, however, we look at shapes Q and R, we can see that theyâ€™re in different orientations.
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Shape R is upside down relative to the position that shape Q was in.
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This means that the transformation weâ€™re looking for is not a translation.
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Next, letâ€™s consider whether or not the transformation weâ€™re looking for could be a rotation.
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A rotation produces an image that is the same size and shape as the original object, but is in a different position and also a different orientation.
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This looks a possibility for the transformation that weâ€™re looking for here.
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To describe a rotation fully, we need to give three further pieces of information.
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We need to give the direction of the rotation, so thatâ€™s clockwise or anticlockwise.
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We need to give the angle of the rotation, how many degrees the shape has been rotated through.
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And we need to give the coordinates of the centre of rotation; this is the point that stays fixed.
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If you were performing a rotation using tracing paper, this is the point where you would stick the point of your pencil and move the tracing paper around about this point.
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The angle is actually the easiest of these three pieces to work out in this case.
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We can see that shape R is completely upside down relative to shape Q, which means itâ€™s been rotated half a turn or 180 degrees.
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Now, this actually means that we donâ€™t need to specify a direction for this rotation because if you turn half a turn one way, youâ€™ll end up in the same place as if you go half a turn the other way.
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So we could answer either clockwise or anticlockwise.
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But actually, we donâ€™t need to give a direction at all.
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Finally, weâ€™ll consider the centre of rotation.
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And as the angle that weâ€™ve rotated through is 180 degrees, the centre of rotation will be somewhere between the two shapes.
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The origin â€” the point zero, zero â€” looks like the most likely candidate.
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But letâ€™s check this.
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If we had access to tracing paper, then we could check this by placing our tracing paper over the diagram, drawing on shape Q, placing our pencil point at the origin and then rotating the tracing paper through 180 degrees or half a turn, and checking that Q ended up directly on top of R.
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But if we donâ€™t have access to tracing paper, letâ€™s look at another method.
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What we do is we draw in horizontal and vertical lines connecting what we think the centre of rotation is â€” so thatâ€™s the origin â€” to each corner on shape Q.
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We then rotate these lines to 180 degrees.
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So the horizontal line on the đť‘Ą-axis going in a positive direction is now a horizontal line going in the negative direction on the đť‘Ą-axis.
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And the vertical line going three units upwards is now a vertical line going three units downwards.
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We can see that this does connect with a corner of shape R.
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We can perform the same check with other points on shape Q and show that they do indeed map to corresponding points on shape R.
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This confirms that the origin or the point zero, zero is indeed the centre of rotation.
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So we can answer that the single transformation that maps the semicircle Q onto the semicircle R is a rotation 180 degrees about the origin or you could say about the point zero, zero.
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Now, the other two types of transformation that we mentioned were reflection and enlargement.
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It isnâ€™t actually possible to place a mirror line anywhere between the two shapes so that shape Q maps directly onto shape R, which means a reflection is not possible.
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However, it is actually possible to map shape Q onto shape R via an enlargement.
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We notice that corresponding lengths on the two shapes are still the same lengths, which means that the shape hasnâ€™t actually been made any bigger.
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Instead, itâ€™s upside down relative to the original shape.
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An enlargement which doesnâ€™t change the size of the original shape, but turns it upside down is an enlargement with a scale factor of negative one.
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This means that all of the corners on the two shapes are the same distance away from the centre of enlargement, but in opposite directions.
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To find the centre of enlargement, we connect corresponding corners on the two shapes together using straight lines.
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Youâ€™ll notice that these three straight lines all crossed at a common point, the origin, which means this is the centre of enlargement.
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There were, therefore, two possible answers for part b.
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Probably, the expected answer is the first one that itâ€™s a rotation of 180 degrees about the origin.
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But if youâ€™re familiar with enlargements with negative scale factors, you could also answer that itâ€™s an enlargement with a scale factor of negative one and a centre of enlargement at the origin.