Intersection and angle between two lines

Two dif­fer­ent lines in a plane either meet at exactly one point or are par­al­lel. This les­son shows how to find the meet­ing point, how to mea­sure the angle between two lines, how to write down at once a line par­al­lel or per­pen­dic­u­lar to a given one, and how to tell whether three lines pass through a sin­gle point.

Point of inter­sec­tion#

The point where two lines meet lies on both, so its coor­di­nates sat­isfy both equa­tions. Solve the two equa­tions simul­ta­ne­ously.

Exam­ple 1#

Find the point of inter­sec­tion of $2x + 3y = 13$ and $x - y = -1$.

From the sec­ond equa­tion, $x = y - 1$. Sub­sti­tute into the first:

$$2(y - 1) + 3y = 13 \implies 5y = 15 \implies y = 3$$

Then $x = 3 - 1 = 2$. The lines meet at $(2, 3)$.

Angle between two lines#

Let two non-ver­ti­cal lines have slopes $m_1$ and $m_2$, and let $\theta$ be the acute angle between them. Then

$$\displaystyle \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$$

Why. If the incli­na­tions are $\theta_1$ and $\theta_2$, the angle between the lines is $\theta_1 - \theta_2$ (or its sup­ple­ment), and

$$\displaystyle \tan(\theta_1 - \theta_2) = \frac{\tan\theta_1 - \tan\theta_2}{1 + \tan\theta_1\tan\theta_2} = \frac{m_1 - m_2}{1 + m_1 m_2}$$

Tak­ing the mod­u­lus gives the acute angle. The for­mula also shows the ear­lier con­di­tions again: the lines are par­al­lel when $m_1 = m_2$ (so $\tan\theta = 0$), and per­pen­dic­u­lar when $1 + m_1 m_2 = 0$ (so $\tan\theta$ is unde­fined and $\theta = 90°$).

Exam­ple 2#

Find the acute angle between $y = 2x + 1$ and $y = -3x + 4$.

Here $m_1 = 2$ and $m_2 = -3$:

$$\displaystyle \tan\theta = \left|\frac{2 - (-3)}{1 + (2)(-3)}\right| = \left|\frac{5}{-5}\right| = 1$$

So $\theta = 45°$.

Exam­ple 3#

Find the acute angle between $x - 3y + 1 = 0$ and $2x - y + 5 = 0$.

The slopes are $\displaystyle m_1 = -\frac{1}{-3} = \frac{1}{3}$ and $\displaystyle m_2 = -\frac{2}{-1} = 2$:

$$\displaystyle \tan\theta = \left|\frac{\frac{1}{3} - 2}{1 + \frac{2}{3}}\right| = \left|\frac{-\frac{5}{3}}{\frac{5}{3}}\right| = 1, \quad \text{so } \theta = 45°$$

Par­al­lel and per­pen­dic­u­lar lines in gen­eral form#

For $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, com­par­ing slopes $\displaystyle -\frac{a_1}{b_1}$ and $\displaystyle -\frac{a_2}{b_2}$ gives:

• Par­al­lel when $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2}$ (that is, $a_1b_2 = a_2b_1$).
• Per­pen­dic­u­lar when $a_1a_2 + b_1b_2 = 0$.

This gives two quick rules for writ­ing down lines.

• A line par­al­lel to $ax + by + c = 0$ has the form $ax + by + k = 0$ (same $x$ and $y$ coef­fi­cients).
• A line per­pen­dic­u­lar to $ax + by + c = 0$ has the form $bx - ay + k = 0$ (swap the coef­fi­cients and change one sign).

The num­ber $k$ is found from a point the line must pass through.

Exam­ple 4#

Find the lines through $(1, -2)$ that are (a) par­al­lel (b) per­pen­dic­u­lar to $3x - 4y + 7 = 0$.

(a) The line is $3x - 4y + k = 0$. Through $(1, -2)$: $3 + 8 + k = 0$, so $k = -11$. The line is $3x - 4y - 11 = 0$.

(b) The line is $4x + 3y + k = 0$. Through $(1, -2)$: $4 - 6 + k = 0$, so $k = 2$. The line is $4x + 3y + 2 = 0$.

Con­cur­rent lines#

Three or more lines are con­cur­rent if they all pass through one point. To test three lines: find where two of them meet, and check whether that point lies on the third.

There is also a deter­mi­nant test: the lines $a_rx + b_ry + c_r = 0$ ($r = 1, 2, 3$), no two of them par­al­lel, are con­cur­rent exactly when

$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$$

Exam­ple 5#

Show that $x + y - 5 = 0$, $2x - y - 1 = 0$ and $3x + 2y - 12 = 0$ are con­cur­rent.

Adding the first two equa­tions: $3x - 6 = 0$, so $x = 2$ and $y = 3$. In the third: $3(2) + 2(3) - 12 = 0$. So all three pass through $(2, 3)$.

By deter­mi­nants: $\begin{vmatrix} 1 & 1 & -5 \\ 2 & -1 & -1 \\ 3 & 2 & -12 \end{vmatrix} = 1(12 + 2) - 1(-24 + 3) - 5(4 + 3) = 14 + 21 - 35 = 0$.

Lines through the meet­ing point of two lines#

Exam­ple 6#

Find the line through the ori­gin and the point of inter­sec­tion of $x + y - 5 = 0$ and $2x - y - 1 = 0$.

From Exam­ple 5 the lines meet at $(2, 3)$. The line through $(0, 0)$ and $(2, 3)$ has slope $\displaystyle \frac{3}{2}$, so it is $\displaystyle y = \frac{3}{2}x$, that is, $3x - 2y = 0$.

Prac­tice#

1. Find the point of inter­sec­tion of $x + 2y = 7$ and $3x - y = 7$.
2. Find the acute angle between $y = 3x + 2$ and $y = -2x + 5$.
3. Find the acute angle between $\sqrt{3}x - y + 1 = 0$ and $x - \sqrt{3}y + 2 = 0$.
4. Show that $2x + 3y - 6 = 0$ and $3x - 2y + 4 = 0$ are per­pen­dic­u­lar.
5. Find the line through $(2, 3)$ par­al­lel to $4x - y + 1 = 0$.
6. Find the line through $(-1, 2)$ per­pen­dic­u­lar to $2x + 5y - 3 = 0$.
7. Find $k$ so that $kx + 2y - 3 = 0$ is per­pen­dic­u­lar to $3x - 4y + 1 = 0$.
8. Show that $x - y + 1 = 0$, $2x + y - 4 = 0$ and $x + 2y - 5 = 0$ are con­cur­rent.
9. Find $k$ so that $x + y - 3 = 0$, $2x - y - 3 = 0$ and $kx + 3y - 7 = 0$ are con­cur­rent.
10. Find the line through the point of inter­sec­tion of $2x + y - 5 = 0$ and $x - y - 1 = 0$, par­al­lel to $3x + 4y = 0$.

Answers#