Video: AQA GCSE Mathematics Higher Tier Pack 2 β€’ Paper 3 β€’ Question 19

A quadrilateral has vertices at the points 𝐴 (1, 1); 𝐡 (1, 4); 𝐢 (5, 5), and 𝐷 (3, 1). It is reflected in a line such that the coordinates of 𝐴 and 𝐢 do not change. Circle the line of reflection. [A] 𝑦 =βˆ’π‘₯ [B] 𝑦 = 1 [C] π‘₯ = 1 [D] 𝑦 = π‘₯.

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Video Transcript

A quadrilateral has vertices at the points 𝐴: one, one; 𝐡: one, four; 𝐢: five, five; and 𝐷: three, one. It is reflected in a line such that the coordinates of 𝐴 and 𝐢 do not change. Circle the line of reflection. The options are 𝑦 equals negative π‘₯, 𝑦 equals one, π‘₯ equals one, or 𝑦 equals π‘₯.

Now it isn’t essential for this question, but we’ll begin with a sketch of this quadrilateral. As all of the vertices have positive π‘₯- and 𝑦-coordinates, the quadrilateral is in the first quadrant. So we can focus our sketch on just this area of the coordinate grid.

Remember, this is just a sketch so that we can visualise the quadrilateral more clearly. We don’t need to plot all of the points accurately.

We’re told that this quadrilateral is reflected in a line such that the coordinates of both 𝐴 and 𝐢 do not change. Points which do not move under transformation are called invariant points. And in the case of a reflection, all invariant points lie along the mirror line itself. This means that, in order to find the line of reflection, the mirror line, we need to work out the equation of the line that joins the points 𝐴 and 𝐢.

Now we could do this formally by using 𝑦 equals π‘šπ‘₯ plus 𝑐, the general form of the equation of a straight line. And we could use the coordinates of points 𝐴 and 𝐢 to calculate the gradient and 𝑦-intercept of this line. However, there are actually a couple of easier ways of doing this.

Firstly, if we look at the coordinates of 𝐴 and 𝐢, we see that, in each case, the π‘₯-coordinate of each point is equal to the 𝑦-coordinate. This will be true for all of the points that lie on the line joining 𝐴 and 𝐢. And if the π‘₯- and 𝑦-coordinates are equal, then the equation of the line that describes this is 𝑦 equals π‘₯.

We could also actually answer this question by process of elimination. We can see from our sketch that the line we’re looking for is a diagonal line of positive gradient. It’s sloping up from left to right. This rules out the possibilities of the lines 𝑦 equals one and π‘₯ equals one, because the line 𝑦 equals one is a horizontal line and the line π‘₯ equals one is a vertical line.

Incidentally, both of these lines do pass through the point 𝐴. So the point 𝐴 would still be invariant under reflection in either of these lines. But as they don’t pass through the point 𝐢, the point 𝐢 would change.

The line 𝑦 equals negative π‘₯ is a diagonal line. But it has a gradient of negative one. So it’s sloping downwards from left to right, which means we can rule this one out too. By process of elimination then, we’d also find that the equation of the line we’re looking for is 𝑦 equals π‘₯.

So you can either formalise the process of finding the equation of this line using 𝑦 equals π‘šπ‘₯ plus 𝑐 or use a sketch diagram or process of elimination to answer this question. The line of reflection which will keep the points 𝐴 and 𝐢 invariant is 𝑦 equals π‘₯.

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