E8.1 - Reflections in oblique lines (Extended)

E8.1 - Reflections in oblique lines (Extended)

A sloping mirror still puts each image directly across a perpendicular, rather than directly left, right, up or down. E8.1 extends reflection to any straight mirror line. You will construct images, join their vertices accurately and use midpoint and right-angle checks to distinguish a reflection from a plausible-looking mistake.

Oblique mirrors

How can you reflect P(4,1)P(4,1) in y=xy=x without guessing? The mirror passes through (1,1)(1,1) and (4,4)(4,4). The reflected point is (1,4)(1,4): its joining segment with P crosses the mirror at right angles and halfway.

Oblique mirror line

An oblique mirror line is sloping rather than horizontal or vertical. The same reflection rule applies: the mirror perpendicularly bisects each point-to-image segment. A point on the mirror stays fixed.

On a squared grid, y=xy=x swaps horizontal and vertical positions, so (x,y)(x,y) maps to (y,x)(y,x). The line y=−xy=-x maps (x,y)(x,y) to (−y,−x)(-y,-x). These are useful special cases, not rules for every sloping line.

For any straight mirror, draw a perpendicular from P to the mirror using a set square or protractor, meeting it at H. Extend PH beyond H. Measure HP′=HPHP'=HP, then mark P'. Use the grid's scale, and ensure the axes use the same physical scale when constructing right angles on paper.

[DIAGRAM: asset_name: Perpendicular constructions in sloping mirrors; asset_slug: oblique_point_construction; caption: H is on the mirror and halfway along the perpendicular joining segment.; description: Two accurate equal-scale grids. Left shows P four comma one reflected in y equals x to P prime one comma four, with H at two point five comma two point five. Right shows P three comma zero reflected in y equals one half x plus one to P prime one comma four, with H at two comma two. Right-angle symbols and matching distance marks establish the construction in both panels.]
H is on the mirror and halfway along the perpendicular joining segment.

Worked example 1: reflection in y equals x

Question. Reflect P(4,1)P(4,1) in y=xy=x.

Plan. Swap coordinates, then check the perpendicular-bisector condition.

P′=(1,4),H=(4+12,1+42)=(2.5,2.5)P'=(1,4),\qquad H=\left(\frac{4+1}{2},\frac{1+4}{2}\right)=(2.5,2.5)

H lies on y=xy=x. From P to P', the move is 3 left and 3 up, perpendicular to a line moving equally right and up.

Answer. P′=(1,4)P'=(1,4).

Check. Repeating the swap returns (4,1)(4,1). The two points lie on opposite sides of the mirror and H is halfway between them.

The next mirror has a different gradient. A coordinate swap would reflect in the wrong line, so use the perpendicular construction.

Worked example 2: a different oblique line

Question. Reflect P(3,0)P(3,0) in y=12x+1y=\frac12x+1, using the right-hand grid.

Plan. Draw PH perpendicular to the mirror and extend the same distance through H.

The perpendicular meets the mirror at H(2,2)H(2,2), as constructed on the grid. From P to H is 1 left and 2 up. Repeat that move from H:

P′=(2−1,2+2)=(1,4)P'=(2-1,2+2)=(1,4)

Answer. P′=(1,4)P'=(1,4).

Check. H satisfies the mirror equation: 2=12(2)+12=\frac12(2)+1. The mirror runs 2 right and 1 up; PH runs 1 left and 2 up, forming a right angle as marked. The identical P-to-H and H-to-P' moves establish equal distances.

To construct a reflection in y=−xy=-x, the same perpendicular-distance method works; the coordinate shortcut merely saves drawing time.

Watch out

Swapping coordinates is valid for y=xy=x, not for every oblique mirror. For example, the swap of (3,0)(3,0) is (0,3)(0,3), which is not its reflection in y=12x+1y=\frac12x+1.

Use the stated mirror in the next task.

Drawing an image

Reflect every vertex separately, label the images and join them with a ruler in the original vertex order. A vertex on an oblique mirror is fixed just as it is on a horizontal mirror. The reflected shape remains congruent to the object.

Problem-solving

For each vertex: draw its perpendicular to the mirror, find the meeting point, then continue by the same distance. Check the vertex labels before joining: joining the right points in the wrong order can make the wrong shape.

The examples below keep the object fixed but change the mirror. A sloping mirror need not pass through the origin.

[DIAGRAM: asset_name: Triangles reflected in two diagonal mirrors; asset_slug: oblique_triangle_images; caption: C lies on each mirror and remains fixed, while A and B move to their corresponding images.; description: Two accurate equal-scale grids show A one comma zero, B three comma zero and C one comma one. The first mirror is y equals x and images are zero comma one, zero comma three and one comma one. The second mirror is y equals minus x plus two and images are two comma one, two comma minus one and one comma one. Object and image edges are ruled, with primes distinguishing corresponding vertices.]
C lies on each mirror and remains fixed, while A and B move to their corresponding images.

Worked example 3: a triangle in a diagonal through the origin

Question. Reflect triangle A(1,0)A(1,0), B(3,0)B(3,0), C(1,1)C(1,1) in y=xy=x.

Plan. Swap the coordinates of every vertex and join the images in order.

A(1,0)↦A′(0,1)B(3,0)↦B′(0,3)C(1,1)↦C′(1,1)\begin{aligned} A(1,0)&\mapsto A'(0,1)\\ B(3,0)&\mapsto B'(0,3)\\ C(1,1)&\mapsto C'(1,1) \end{aligned}

Answer. The image is the triangle joining (0,1),(0,3),(1,1)(0,1),(0,3),(1,1), labelled A', B', C' as on the left grid.

Check. C lies on y=xy=x, so stays fixed. AB is horizontal with length 2; A'B' is vertical with length 2. Every midpoint of a moved pair lies on y=xy=x.

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