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Secondary 5 · Mathematics
Vectors and geometry · Lines in the plane

Angle Between Two Lines. Parallelism and Perpendicularity

In Santo Domingo, an engineer designs a new avenue that must cross the Autopista Duarte at an exact angle. On a coordinate plane you don't need a protractor: the equations of the two lines are enough to calculate the angle. How is it done, and how can you make sure two streets are parallel or perpendicular?

The Angle Between Two Lines

Two intersecting lines form four angles. If they are not perpendicular, two of the angles are acute and equal, and the other two are obtuse and equal. If they are perpendicular, all four measure $90^\circ$. By convention, the angle between two lines is the acute one (between $0^\circ$ and $90^\circ$), or the right angle if they are perpendicular.

This formula comes from the difference of two angles of inclination: if $m_1=\tan\theta_1$ and $m_2=\tan\theta_2$, then $\alpha=\theta_1-\theta_2$ and the tangent of a difference is used.

Two intersecting lines r and s that meet at the origin forming an acute angle alpha in the first quadrant, marked with an arc between them. α r s

If the line is given only by its slope, remember that $m=\tan\theta$, where $\theta$ is the angle it forms with the positive direction of the $x$-axis. So a line with slope $\sqrt{3}$ forms $60^\circ$ with the $x$-axis.

Practice★☆☆Not started
Calculate the angle formed by the lines given by their explicit equations $y = 2x + 1$ and $y = -3x + 4$.

Calculate the angle formed by the lines given by their explicit equations $y = 2x + 1$ and $y = -3x + 4$.

Solve
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Calculate the acute angle between the lines $r: 2x - y + 4 = 0$ and $s: 3x + y - 1 = 0$.

Calculate the acute angle between the lines $r: 2x - y + 4 = 0$ and $s: 3x + y - 1 = 0$.

Solve
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Determine the angle that the line $\sqrt{3}x - y + 2 = 0$ forms with the positive direction of the x-axis.

Determine the angle that the line $\sqrt{3}x - y + 2 = 0$ forms with the positive direction of the x-axis.

Solve

Parallel Lines

Why? The slope is the tangent of the angle of inclination. If two lines are tilted the same way, they go in the same direction and never meet. In the angle formula, $m_1=m_2$ gives $\tan\alpha=0$, that is, $\alpha=0^\circ$.

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Determine the general equation of the line parallel to $y = 3x - 5$ passing through the point $P(2, -1)$.

Determine the general equation of the line parallel to $y = 3x - 5$ passing through the point $P(2, -1)$.

Solve
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Given the lines $r: kx - 2y + 5 = 0$ and $s: 3x + 4y - 1 = 0$, find the value of the parameter $k$ so that both lines are parallel.

Given the lines $r: kx - 2y + 5 = 0$ and $s: 3x + 4y - 1 = 0$, find the value of the parameter $k$ so that both lines are parallel.

Solve

Perpendicular Lines

This agrees with the angle formula: $1+m_1m_2=0$ makes $\alpha=90^\circ$. If a line is horizontal ($m=0$), its perpendicular is vertical and has no slope.

Two perpendicular lines in the Cartesian plane that intersect forming a right angle marked with a small square.
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Find the equation of the line perpendicular to $x + 4y - 7 = 0$ passing through the origin of coordinates. Express the result in explicit form.

Find the equation of the line perpendicular to $x + 4y - 7 = 0$ passing through the origin of coordinates. Express the result in explicit form.

Solve
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Find the value of $m$ so that the lines $r: 5x + 2y = 0$ and $s: mx - 5y + 3 = 0$ are perpendicular.

Find the value of $m$ so that the lines $r: 5x + 2y = 0$ and $s: mx - 5y + 3 = 0$ are perpendicular.

Solve
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State which of the following lines are parallel and which are perpendicular to each other: $r_1: 2x - 3y + 1 = 0$ $r_2: 6x + 4y - 5 = 0$ $r_3: -4x + 6y + 7 = 0$

State which of the following lines are parallel and which are perpendicular to each other:

  • $r_1: 2x - 3y + 1 = 0$

  • $r_2: 6x + 4y - 5 = 0$

  • $r_3: -4x + 6y + 7 = 0$

Solve

Orthogonal Projection of a Point onto a Line

The orthogonal projection of a point $P$ onto a line $r$ is the point $H$ on $r$ closest to $P$: the foot of the perpendicular drawn from $P$.

Line r, a point P off the line, and its orthogonal projection H onto r, with the perpendicular PH and the right-angle mark. P H r

It is found in three steps: perpendicular slope, perpendicular line through $P$, and intersection with $r$ (a 2×2 system).

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Find the coordinates of the orthogonal projection of the origin of coordinates $(0,0)$ onto the line with equation $2x - 5y + 10 = 0$.

Find the coordinates of the orthogonal projection of the origin of coordinates $(0,0)$ onto the line with equation $2x - 5y + 10 = 0$.

Solve

Applications: Altitudes, Perpendicular Bisectors and Structures

Perpendicularity shows up whenever something must meet at a right angle: the altitude of a triangle, the perpendicular bisector of a segment (perpendicular through the midpoint), a crossbeam in a structure, or a new street that crosses another. The diagonals of a square or a rhombus are also perpendicular, and they cross at their midpoint.

Triangle ABC with the altitude drawn from vertex C perpendicular to base AB, with a right-angle mark. A B C altitude
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In the design of a park, a new avenue must be perpendicular to the main street defined by the equation $3x - 4y + 10 = 0$. Furthermore, this avenue must pass through the…

In the design of a park, a new avenue must be perpendicular to the main street defined by the equation $3x - 4y + 10 = 0$. Furthermore, this avenue must pass through the point of intersection of the side streets $x + y - 2 = 0$ and $2x - y - 4 = 0$. Obtain the general equation of the new avenue.

Solve
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Given a triangle with vertices $A(1, 2)$, $B(5, 3)$ and $C(4, 6)$, find the general equation of the altitude corresponding to vertex $A$ and calculate the angle this…

Given a triangle with vertices $A(1, 2)$, $B(5, 3)$ and $C(4, 6)$, find the general equation of the altitude corresponding to vertex $A$ and calculate the angle this altitude forms with the side $AB$.

Solve
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In the plan for a metallic structure, a main support beam rests on the line passing through nodes $P(-2, 4)$ and $Q(4, 1)$. A second beam needs to be welded…

In the plan for a metallic structure, a main support beam rests on the line passing through nodes $P(-2, 4)$ and $Q(4, 1)$. A second beam needs to be welded perpendicular to the first exactly at its midpoint. Find the explicit equation of the line defining this second beam.

Solve

Reflection: Reflected Ray and Symmetric Line

When a ray of light hits a flat mirror, the angle it forms with the normal (the perpendicular to the mirror at the point of impact) equals the angle of the reflected ray. This is the angle of incidence.

An incident ray hits a horizontal mirror, is reflected, and forms the same angle with the vertical normal as the reflected ray. incident reflected normal mirror

To reflect a line $r$ across another line $s$ (the symmetric line): find the point $P$ where they intersect, reflect a point of $r$ across $s$ (perpendicular through that point, foot of the perpendicular, and double the distance), and join the reflected point to $P$.

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A laser beam travels along the line $y = -2x + 5$ and reflects off a flat mirror situated on the x-axis. Calculate the angle of incidence (angle between the beam and the…

A laser beam travels along the line $y = -2x + 5$ and reflects off a flat mirror situated on the x-axis. Calculate the angle of incidence (angle between the beam and the normal to the mirror) and determine the equation of the line containing the reflected beam.

X Y Mirror Incident Reflected Normal α α
Solve
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Determine the general equation of the line symmetric to $r: 2x - 3y + 6 = 0$ with respect to the line $s: x + y - 1 = 0$.

Determine the general equation of the line symmetric to $r: 2x - 3y + 6 = 0$ with respect to the line $s: x + y - 1 = 0$.

Solve

Extension: Lines with a Parameter

When a line depends on a parameter $k$, the required condition becomes an equation for $k$: parallelism ($m_1=m_2$), perpendicularity ($m_1m_2=-1$), or a given angle ($\tan\alpha$ equal to the formula).

The same method works to find lines that form a given angle with a known line: set up $\tan\alpha=\left|\dfrac{m-m_1}{1+mm_1}\right|$ with the slope $m$ as the unknown. For example, an equilateral triangle forces sides that form $60^\circ$ ($\tan60^\circ=\sqrt3$) with another side, and two lines result.

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Calculate the values of the parameter $k$ so that the line $kx + y - 2 = 0$ forms an angle of $45^\circ$ with the line with equation $x - 2y + 1 = 0$.

Calculate the values of the parameter $k$ so that the line $kx + y - 2 = 0$ forms an angle of $45^\circ$ with the line with equation $x - 2y + 1 = 0$.

Solve
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Two aeroplanes follow straight-line paths on a radar screen given by the equations $r: x - 3y + 2 = 0$ and $s: 2x + ky - 5 = 0$. Determine the value (or values) of $k$…

Two aeroplanes follow straight-line paths on a radar screen given by the equations $r: x - 3y + 2 = 0$ and $s: 2x + ky - 5 = 0$. Determine the value (or values) of $k$ so that their routes cross, forming an exact angle of $60^\circ$.

Solve
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Two robotic arms in a factory move along straight rails. The first follows the line $r: 3x + 4y - 12 = 0$ and the second the line $s: ax + 8y - 5 = 0$. 1. Determine $a$…

Two robotic arms in a factory move along straight rails. The first follows the line $r: 3x + 4y - 12 = 0$ and the second the line $s: ax + 8y - 5 = 0$.

  1. Determine $a$ so that the rails are parallel.

  2. Find the equation of a third transverse rail that is perpendicular to both and crosses rail $r$ exactly at its y-intercept.

Solve

Extension: Families of Lines with a Fixed Point

An equation with a parameter, such as $(1+k)x+(2-k)y-3=0$, describes a family of lines: a different line for each value of $k$. Sometimes all of them pass through the same point.

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Given the family of lines depending on the parameter $k \in \mathbb{R}$: $(2 + k)x + (1 - 2k)y + 3k - 4 = 0$. Show that all the lines in this family pass through the…

Given the family of lines depending on the parameter $k \in \mathbb{R}$: $(2 + k)x + (1 - 2k)y + 3k - 4 = 0$. Show that all the lines in this family pass through the same fixed point and calculate its coordinates. Then, find the value of $k$ to obtain the line in the family that is perpendicular to $3x - y + 2 = 0$.

Solve

Extension: Angle Bisectors of Two Lines

Two intersecting lines form four angles and have two angle bisectors. Every point on a bisector is the same distance from the two lines.

This always happens: the two bisectors are perpendicular to each other, because adjacent angles add up to $180^\circ$ and each bisector splits one of them in half (the sum of the two halves is $90^\circ$).

Practice★★★Not started
Find the equations of the angle bisectors of the angles formed by the lines $r: 3x - 4y + 5 = 0$ and $s: 12x + 5y - 1 = 0$. Prove analytically, using their direction…

Find the equations of the angle bisectors of the angles formed by the lines $r: 3x - 4y + 5 = 0$ and $s: 12x + 5y - 1 = 0$. Prove analytically, using their direction vectors or their gradients, that both bisectors are perpendicular to each other.

Solve

Summary

Go back to the beginning: for the new avenue to cross at a right angle, it is enough for the product of the slopes of the two streets to be $-1$.

See all the exercises in this section