You don't know what the lines are, but can you find the angle between them?

Geometry Level 5

A point $P$ , starting at $(2,6)$ , was reflected across line $m$ in the $(x,y)$ plane to produce point $P^\prime$ .

$P^\prime$ was then reflected across line $n$ in the $(x,y)$ plane to produce point $P^{\prime\prime}$ at $(6,4)$ .

If the lines $m$ and $n$ intersect at the point $(2,1)$ , what is the angle, in degrees, that these lines form at their intersection?

Your answer should be less than 90 degrees, and correct to 1 decimal place.

The answer is 26.6.

2 solutions

Worranat Pakornrat
Nov 9, 2015

Point (2 , 1) is actually the center O of the circle with radius 5, intersecting (2 , 6) & (6 , 4).

If we draw lines m and n to cause reflection, these lines will act as bisectors between PP' and P'P'' respectively.

That is triangle POP' & P'OP'' are both isoscales triangles with equal sides of 5. As a result, the angle POP'' will be twice as much as the angle made from lines m & n because they are the bisectors.

Now considering triangle POP'', it's also an isoscales triangle, and PP'' = sq.rt.((2-6)^2+(6-4)^2) = 2(sq.rt.(5)).

Then if we draw a bisector line halving PP" at midpoint Q. PQO is a right triangle, and angle QOP is the desired angle for solution. (Angle QOP = angle POP"/2)

Then sin QOP = PQ/OP = sq.rt.(5)/5 = 1/sq.rt.(5).

Finally, sinh(1/sq.rt.(5)) is about 26.6 degrees, which is the angle line m & n make.

M M
Nov 6, 2015

The transformation composition of reflecting over two intersecting lines produces a rotation around the point of intersection. (This fact is occasionally taught in geometry class, but a proof will be forthcoming in the transformations section of this site as my time allows...). The resulting angle of rotation is twice the angle formed between the two lines of reflection.

There are multiple ways to determine the angle of rotation, which I will call $\theta$ , from point P to P'' around the center of rotation at (2,1), which I will call C. You can draw a 3-4-5 right triangle with one leg of length 3 along the segment CP, and the hypotenuse being CP". Then

$cos(\theta) = 3/5$

$\Rightarrow \theta\approx53.13°$

Since the angle of rotation is twice the angle between the two lines of reflection, the angle we seek is

$53.13 / 2 \approx 26.6°$

A more general solution to this problem, however, uses vector dot products to determine the angle PCP".

The vector CP = <0,5> and the vector CP" = <4,3>

Using the formula $cos(\theta)=\frac{CP \cdot CP"}{\| CP\| \| CP"\|}$ ,

$cos(\theta)=\frac{(0\cdot 4) + (5\cdot 3)}{5\cdot 5} = \frac{15}{25} = \frac{3}{5}$ , which provides us the same answer as before.

You don't know what the lines are, but can you find the angle between them?

The answer is 26.6.

2 solutions

0 pending reports