Foot of the perpendicular, image and centres of a triangle

The last les­son found how far a point is from a line. This one finds where: the foot of the per­pen­dic­u­lar from the point to the line, and the mir­ror image (reflec­tion) of the point in the line. It ends by using per­pen­dic­u­lar lines to locate two spe­cial points of a tri­an­gle, the ortho­cen­tre and the cir­cum­cen­tre, bring­ing together the tools of the whole course.

Foot of the per­pen­dic­u­lar#

Result. If $(h, k)$ is the foot of the per­pen­dic­u­lar from $P(x_1, y_1)$ to the line $ax + by + c = 0$, then

$$\displaystyle \frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$$

Why. The per­pen­dic­u­lar from $P$ has slope $\displaystyle \frac{b}{a}$ (per­pen­dic­u­lar to slope $\displaystyle -\frac{a}{b}$), so its points can be writ­ten $(x_1 + at, y_1 + bt)$ for some num­ber $t$. The foot is the one of these that lies on the line:

$$\displaystyle a(x_1 + at) + b(y_1 + bt) + c = 0 \implies t = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$$

and $h = x_1 + at$, $k = y_1 + bt$ give the result.

Exam­ple 1#

Find the foot of the per­pen­dic­u­lar from $(3, 4)$ to $x + y - 1 = 0$.

Here $a = b = 1$, $c = -1$:

$$\displaystyle \frac{h - 3}{1} = \frac{k - 4}{1} = -\frac{3 + 4 - 1}{1 + 1} = -3$$

So $h = 3 - 3 = 0$ and $k = 4 - 3 = 1$. The foot is $(0, 1)$. Check: $0 + 1 - 1 = 0$, so it lies on the line.

Exam­ple 2#

Find the foot of the per­pen­dic­u­lar from $(1, 2)$ to $3x - 4y + 15 = 0$.

$$\displaystyle \frac{h - 1}{3} = \frac{k - 2}{-4} = -\frac{3 - 8 + 15}{9 + 16} = -\frac{10}{25} = -\frac{2}{5}$$

So $\displaystyle h = 1 - \frac{6}{5} = -\frac{1}{5}$ and $\displaystyle k = 2 + \frac{8}{5} = \frac{18}{5}$. The foot is $\displaystyle \left(-\frac{1}{5}, \frac{18}{5}\right)$.

As a check, the dis­tance from $(1, 2)$ to the foot is $\displaystyle \sqrt{\left(\frac{6}{5}\right)^2 + \left(\frac{8}{5}\right)^2} = \sqrt{\frac{100}{25}} = 2$, which equals $\displaystyle \frac{|3 - 8 + 15|}{5} = 2$ from the dis­tance for­mula.

Image of a point in a line#

The image (reflec­tion) $Q$ of a point $P$ in a line is the point on the other side such that the line is the per­pen­dic­u­lar bisec­tor of $PQ$. So the foot of the per­pen­dic­u­lar is the mid­point of $P$ and its image, and the image is twice as far along the per­pen­dic­u­lar.

Result. If $(h, k)$ is the image of $P(x_1, y_1)$ in $ax + by + c = 0$, then

$$\displaystyle \frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{2(ax_1 + by_1 + c)}{a^2 + b^2}$$

Exam­ple 3#

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

$$\displaystyle \frac{h - 3}{1} = \frac{k - 4}{1} = -\frac{2(6)}{2} = -6$$

So the image is $(3 - 6, 4 - 6) = (-3, -2)$. Check: the mid­point of $(3, 4)$ and $(-3, -2)$ is $(0, 1)$, the foot found in Exam­ple 1.

Images in spe­cial lines. The image of $(x, y)$

• in the $x$-axis is $(x, -y)$;
• in the $y$-axis is $(-x, y)$;
• in the ori­gin is $(-x, -y)$;
• in the line $y = x$ is $(y, x)$.

Cen­tres of a tri­an­gle#

Exam­ple 4#

Find the ortho­cen­tre of the tri­an­gle with ver­tices $A(0, 0)$, $B(6, 0)$, $C(2, 4)$.

The ortho­cen­tre is the point where the three alti­tudes meet. An alti­tude passes through a ver­tex, per­pen­dic­u­lar to the oppo­site side. Two alti­tudes are enough.

• Alti­tude from $C$: $AB$ lies along the $x$-axis, so the alti­tude is the ver­ti­cal line $x = 2$.
• Alti­tude from $A$: the slope of $BC$ is $\displaystyle \frac{4 - 0}{2 - 6} = -1$, so the alti­tude has slope $1$ and passes through $A$: $y = x$.

These meet at $(2, 2)$. Check with the third alti­tude, from $B$: the slope of $AC$ is $2$, so the alti­tude is $\displaystyle y - 0 = -\frac{1}{2}(x - 6)$, which at $x = 2$ gives $y = 2$. The ortho­cen­tre is $(2, 2)$.

Exam­ple 5#

Find the cir­cum­cen­tre of the same tri­an­gle.

The cir­cum­cen­tre is the point where the per­pen­dic­u­lar bisec­tors of the sides meet; it is equally dis­tant from all three ver­tices.

• Per­pen­dic­u­lar bisec­tor of $AB$: the mid­point is $(3, 0)$ and $AB$ is hor­i­zon­tal, so the bisec­tor is $x = 3$.
• Per­pen­dic­u­lar bisec­tor of $AC$: the mid­point is $(1, 2)$ and the slope of $AC$ is $2$, so the bisec­tor is $\displaystyle y - 2 = -\frac{1}{2}(x - 1)$.

At $x = 3$: $y = 2 - 1 = 1$. The cir­cum­cen­tre is $(3, 1)$. Check: its dis­tances from $A$, $B$, $C$ are $\sqrt{9 + 1}$, $\sqrt{9 + 1}$ and $\sqrt{1 + 9}$, all equal to $\sqrt{10}$.

Prac­tice#

1. Find the foot of the per­pen­dic­u­lar from $(2, 5)$ to $x - y + 1 = 0$.
2. Find the image of $(2, 5)$ in the line $x - y + 1 = 0$.
3. Find the foot of the per­pen­dic­u­lar from the ori­gin to $3x + 4y - 25 = 0$.
4. Find the image of $(1, 2)$ in the line $3x - 4y + 15 = 0$.
5. Find the images of $(4, -3)$ in the $x$-axis, the $y$-axis and the line $y = x$.
6. Find the image of $(-2, 5)$ in the line $x + 2y = 0$.
7. Find the ortho­cen­tre of the tri­an­gle with ver­tices $(0, 0)$, $(4, 0)$, $(0, 6)$.
8. Find the cir­cum­cen­tre of the tri­an­gle with ver­tices $(0, 0)$, $(4, 0)$, $(0, 6)$.
9. Find the ortho­cen­tre of the tri­an­gle with ver­tices $A(1, 1)$, $B(5, 1)$, $C(3, 5)$.
10. Find the cir­cum­cen­tre of the tri­an­gle in ques­tion 9.

Answers#