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4.3 C) Reflection: Finding the Mirror Line
4.3 C) Reflection: Finding the Mirror Line
In this section we are going to be looking at a few questions whereby we are given the shape and the image and are asked to give an equation for the mirror line. The best way of explaining the process to answer these types of questions is to have an example.
Example 1
The shape (A) has been reflected to create the image (B). Find an equation for the mirror line.
The shape (A) has been reflected to create the image (B). Find an equation for the mirror line.
From looking at the above diagram, we can see that the mirror line is going to be a vertical line. The first step in finding the mirror line is to make a mark half away between the respective points on the shape and the image. I am going to do this for the top two points, and I have drawn a line connecting the top two points on the diagram below to help us work out the distances.
The distance between the top point on the shape and the image is 4 squares. Half of 4 is 2, so we make a mark on the diagram 2 squares from each of the points. This mark is shown on the diagram below.
We do the same for the other two points and the outcome is shown below.
We now have 3 points that lie on the mirror line. The final step is to connect these points together to create a line. The mirror line has been drawn on the graph below.
We now have the mirror line, but the question has asked for an equation of the mirror line. The line above is a vertical line and this means that the equation will take the form x = “something”. The something in this case will be 1. Therefore, the equation of the mirror line is x = 1.
Example 2
The shape (blue) has been reflected in a line to produce an image (green). Find an equation for the mirror line.
The shape (blue) has been reflected in a line to produce an image (green). Find an equation for the mirror line.
This question is slightly different to the previous one because we are given the points on the shape and their respective points on the image. This means that we are able to move onto the next step, which is to draw lines going from the points on the shape to their respective points on the image. We draw lines going from A to A’, B to B’ and C to C’.
Half of the distance between each of the respective points will lie on the mirror line. Let’s work this out for point A and A’. The distance between point A and A’ is 2 diagonal squares. Half of 2 diagonal squares is 1 diagonal square, which means that the point that lies on the mirror line will be 1 diagonal square from each of the points (point A and A’). This point is shown on the diagram below.
We do the same for the other two points on the shape. The midpoints for the other two respective points are shown on the graph below.
The next step is to draw a line that connects all of the midpoints between the points on the shape and their respective point on the image. The line that is drawn will be the mirror line.
We now need to find an equation for the mirror line. This line is a downwards sloping line, which means that it will take the following form:
m in the above formula is the gradient of the line and c is the y intercept. The gradient is the slope of the line. The above line is downwards sloping and as we increase the x value by 1, the y value decreases by 1. This means that the gradient of the line is -1; m is -1. We now need to find the y intercept, which is the y coordinates where the line crosses the y-axis. The y intercept for the above graph is 3, which means that c is 3. We can combine these two to give us the following equation:
This equation is the same as y = -1x + 3. Whenever we have 1 as the coefficient of an unknown, there is no need to include the 1.