Angle bisectors and reflections

Two bisectors from one equation

Two lines that meet make four angles, in two pairs. The two bisectors cross at right angles. Every point on a bisector is equally far from both lines, and every point equally far from both lines is on a bisector.

Take $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$. Equal distances mean

The two signs give the two bisectors.

Example 1. Find the bisectors of $5x + 12y + 7 = 0$ and $12x + 5y + 21 = 0$.

Both lines have $\sqrt{25 + 144} = 13$ underneath, so the denominators cancel.

• Plus sign. $5x + 12y + 7 = 12x + 5y + 21$ gives $-7x + 7y - 14 = 0$, which is $x - y + 2 = 0$.
• Minus sign. $5x + 12y + 7 = -(12x + 5y + 21)$ gives $17x + 17y + 28 = 0$.

Check: the slopes are $1$ and $-1$, so the two bisectors are at right angles.

For two lines through the origin at angles $\alpha$ and $\beta$ to the $x$-axis, the bisectors are at $\dfrac{\alpha + \beta}{2}$ and $\dfrac{\alpha + \beta}{2} + 90^\circ$. The slope of the first is $\tan\dfrac{\alpha + \beta}{2}$, not the average of the two slopes.

Which bisector is which?

Problems ask for the bisector of the acute angle, or of the angle that holds a given point. Use this rule.

1. Make both constants positive. Multiply a line by $-1$ if needed, so that $c_1 > 0$ and $c_2 > 0$.
2. The plus sign goes through the origin's angle. The bisector with the plus sign passes through the angle that contains the origin.
3. Read the angle from the signs of $a_1a_2 + b_1b_2$. If it is negative, the origin's angle is acute. If it is positive, the origin's angle is obtuse. The other bisector bisects the other angle.

Here is why. With $c > 0$, the origin lies on the side of each line where the expression is positive. The angle that holds the origin lies between two half-planes, with normals $(a_1, b_1)$ and $(a_2, b_2)$. When the normals make an acute angle, the angle between the lines that holds the origin is obtuse, and the other way round.

In Example 1, $c_1 = 7$ and $c_2 = 21$ are positive, and $a_1a_2 + b_1b_2 = 5 \times 12 + 12 \times 5$, which is positive. So the origin's angle is obtuse, about 135 degrees, and $x - y + 2 = 0$ bisects it. The acute-angle bisector is $17x + 17y + 28 = 0$.

If one line passes through the origin, so that $c = 0$, test with another point instead. A point is in an angle when it gives the same signs as the points inside that angle.

The image of a point in a line

The mirror image $P'$ of $P(x_1, y_1)$ in the line $ax + by + c = 0$ lies on the perpendicular from $P$, as far behind the line as $P$ is in front. Let

The foot of the perpendicular is $(x_1 - at,\ y_1 - bt)$, half way to the image. The image itself is

Example 2. Find the image of $(4, 3)$ in the line $x + 2y = 15$.

Here $t = \dfrac{4 + 6 - 15}{1 + 4} = -1$. The foot is $(4 + 1,\ 3 + 2) = (5, 5)$ and the image is $(6, 7)$.

Check. The middle point of $(4, 3)$ and $(6, 7)$ is $(5, 5)$, which lies on the line since $5 + 10 = 15$. The segment between them has slope $2$, and the mirror has slope $-\dfrac{1}{2}$. The product is $-1$.

To reflect a whole line, keep the point where it meets the mirror, which does not move, and reflect one other point of the line. Join the two.

The path of a light ray

A ray leaves $A$, strikes a mirror, and arrives at $B$, with $A$ and $B$ on the same side. Among all points of the mirror, the one where the ray strikes makes the trip $A \to M \to B$ as short as possible.

Reflect $B$ in the mirror to get $B'$. The straight line $AB'$ meets the mirror at $M$. The trip $A \to M \to B$ is as long as the line $AB'$, and any other point of the mirror gives a longer trip.

Example 3. A ray goes from $A(0, 2)$ to $B(5, 1)$ by way of the mirror $x + y = 10$.

1. Reflect $B$. $t = \dfrac{5 + 1 - 10}{2} = -2$, so $B' = (5 + 4,\ 1 + 4) = (9, 5)$.
2. Join $A$ to $B'$. The slope is $\dfrac{5 - 2}{9 - 0} = \dfrac{1}{3}$, so the line is $y = 2 + \dfrac{x}{3}$.
3. Meet the mirror. $x + 2 + \dfrac{x}{3} = 10$ gives $x = 6$ and $y = 4$. The ray strikes at $M(6, 4)$.

The length of the whole trip is $AB' = \sqrt{81 + 9} = \sqrt{90}$.

Practise with new numbers

Try it in Python

The program finds both bisectors of two lines, checks that every point on them is equally far from both lines, and counts the angle that holds the origin.

Common mistakes

• Forgetting the second bisector. A modulus gives two lines. A question about the acute-angle bisector says which one it wants.
• Mixing the signs of the constants. The rule about the origin works only when both constants are positive. Multiply a line by $-1$ first if needed.
• Skipping the denominators. Divide each line by its own $\sqrt{a^2 + b^2}$ before you equate, unless the two are equal.
• Stopping at the foot. The foot of the perpendicular is half way to the image. Go as far again.

Remember these

• The bisectors are $\dfrac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm\dfrac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$.
• With $c_1, c_2 > 0$, the plus sign goes through the origin's angle, and that angle is acute when $a_1a_2 + b_1b_2 < 0$.
• The image of $(x_1, y_1)$ is $(x_1 - 2at,\ y_1 - 2bt)$ with $t = \dfrac{ax_1 + by_1 + c}{a^2 + b^2}$.
• For a light ray, reflect one point, join, and meet the mirror.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the bisector of the acute angle between $3x + 4y = 7$ and $4x + 3y = 7$.

2. Find the image of $(4, 1)$ in the line $2x - y + 3 = 0$.

3. A ray goes from $(0, 3)$ to the $x$-axis and then through $(8, 5)$. Where does it strike the axis?

4. Two lines through the origin make angles of 25 and 85 degrees with the $x$-axis. What angle does the bisector of the acute angle between them make with the $x$-axis?