Intersection and Angles Between Lines

1. Point of Intersection of Two Lines

When two distinct lines in a plane are not parallel, they must intersect at exactly one point. Finding this point is a fundamental problem in coordinate geometry. We can find the point of intersection by solving the system of linear equations that represent the two lines simultaneously.

Consider two lines with equations:

Line 1: a₁x + b₁y + c₁ = 0

Line 2: a₂x + b₂y + c₂ = 0

To find the point of intersection (x, y), we solve these two equations. There are several methods to do this, including substitution, elimination, and using determinants (Cramer's Rule).

Method 1: Elimination Method

Multiply the first equation by b₂ and the second by b₁:

a₁b₂x + b₁b₂y + c₁b₂ = 0

a₂b₁x + b₁b₂y + c₂b₁ = 0

Subtract the second modified equation from the first:

(a₁b₂ - a₂b₁)x + (c₁b₂ - c₂b₁) = 0

If (a₁b₂ - a₂b₁) ≠ 0, then:

x = (c₂b₁ - c₁b₂) / (a₁b₂ - a₂b₁)

Similarly, multiply the first equation by a₂ and the second by a₁:

a₁a₂x + b₁a₂y + c₁a₂ = 0

a₁a₂x + b₂a₁y + c₂a₁ = 0

Subtract the first modified equation from the second:

(b₂a₁ - b₁a₂)y + (c₂a₁ - c₁a₂) = 0

If (a₁b₂ - a₂b₁) ≠ 0, then:

y = (c₁a₂ - c₂a₁) / (a₁b₂ - a₂b₁)

Method 2: Cramer's Rule

For the system of equations:

a₁x + b₁y = -c₁

a₂x + b₂y = -c₂

The determinant of the coefficient matrix is D = a₁b₂ - a₂b₁.

The determinant D_x is obtained by replacing the x-coefficients with the constants:

D_x = (-c₁)b₂ - (-c₂)b₁ = c₂b₁ - c₁b₂

The determinant D_y is obtained by replacing the y-coefficients with the constants:

D_y = a₁(-c₂) - a₂(-c₁) = c₁a₂ - c₂a₁

If D ≠ 0, the unique solution is:

x = D_x / D = (c₂b₁ - c₁b₂) / (a₁b₂ - a₂b₁)

y = D_y / D = (c₁a₂ - c₂a₁) / (a₁b₂ - a₂b₁)

Example:

Find the point of intersection of the lines 2x + 3y - 5 = 0 and x - y + 1 = 0.

Here, a₁=2, b₁=3, c₁=-5 and a₂=1, b₂=-1, c₂=1.

Using Cramer's Rule:

D = (2)(-1) - (1)(3) = -2 - 3 = -5

D_x = (1)(3) - (-5)(-1) = 3 - 5 = -2

D_y = (-5)(1) - (1)(2) = -5 - 2 = -7

x = D_x / D = -2 / -5 = 2/5

y = D_y / D = -7 / -5 = 7/5

The point of intersection is (2/5, 7/5).

2. Angle Between Two Lines

The angle between two intersecting lines is the acute angle formed at their intersection point. If the lines have slopes m₁ and m₂, the tangent of the angle θ between them is given by the formula:

tan θ = |(m₁ - m₂) / (1 + m₁m₂)|

This formula gives the tangent of the acute angle. If (1 + m₁m₂) = 0, then m₁m₂ = -1, which means the lines are perpendicular, and the angle is 90 degrees (π/2 radians).

Derivation of the formula:

Let α₁ and α₂ be the angles of inclination of the two lines with the positive x-axis. So, m₁ = tan α₁ and m₂ = tan α₂.

The angle θ between the two lines is the difference between their angles of inclination, i.e., θ = α₁ - α₂ or θ = α₂ - α₁.

Using the tangent subtraction formula:

tan θ = tan(α₁ - α₂) = (tan α₁ - tan α₂) / (1 + tan α₁ tan α₂)

Substituting m₁ and m₂:

tan θ = (m₁ - m₂) / (1 + m₁m₂)

Since we are usually interested in the acute angle, we take the absolute value:

tan θ = |(m₁ - m₂) / (1 + m₁m₂)|

Special Cases:

1. Perpendicular Lines: If two lines are perpendicular, the product of their slopes is -1. That is, m₁m₂ = -1. In this case, the denominator (1 + m₁m₂) becomes zero, and tan θ approaches infinity, meaning θ = 90°.
2. Parallel Lines: If two lines are parallel, their slopes are equal, m₁ = m₂. In this case, the numerator (m₁ - m₂) becomes zero, so tan θ = 0, meaning θ = 0°.

Angle between lines in general form:

Consider two lines given in the general form:

Line 1: a₁x + b₁y + c₁ = 0. Its slope is m₁ = -a₁/b₁.

Line 2: a₂x + b₂y + c₂ = 0. Its slope is m₂ = -a₂/b₂.

Substitute these into the formula:

tan θ = |((-a₁/b₁) - (-a₂/b₂)) / (1 + (-a₁/b₁)(-a₂/b₂))|

tan θ = |((-a₁b₂ + a₂b₁) / b₁b₂) / ((b₁b₂ + a₁a₂) / b₁b₂)|

tan θ = |(a₂b₁ - a₁b₂) / (a₁a₂ + b₁b₂)|

Alternatively, we can use the dot product of their normal vectors. The normal vector to Line 1 is n₁ = (a₁, b₁) and to Line 2 is n₂ = (a₂, b₂).

The angle φ between the normal vectors is given by:

cos φ = (n₁ · n₂) / (|n₁| |n₂|) = (a₁a₂ + b₁b₂) / (√(a₁² + b₁²) √(a₂² + b₂²))

The angle between the lines is either φ or 180° - φ. The acute angle θ between the lines is such that cos θ = |cos φ|.

So, cos θ = |(a₁a₂ + b₁b₂)| / (√(a₁² + b₁²) √(a₂² + b₂²))

And tan θ = |(a₂b₁ - a₁b₂)| / (a₁a₂ + b₁b₂).

Example:

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

Line 1: -2x + y + 1 = 0. Slope m₁ = -(-2)/1 = 2.

Line 2: 3x + y - 2 = 0. Slope m₂ = -(3)/1 = -3.

Using the formula tan θ = |(m₁ - m₂) / (1 + m₁m₂)|:

tan θ = |(2 - (-3)) / (1 + (2)(-3))|

tan θ = |(2 + 3) / (1 - 6)|

tan θ = |5 / -5| = |-1| = 1

Since tan θ = 1, the acute angle θ = 45° or π/4 radians.

Example 2:

Find the angle between the lines 3x + 4y - 7 = 0 and 5x - 12y + 3 = 0.

Line 1: a₁=3, b₁=4.

Line 2: a₂=5, b₂=-12.

Using cos θ = |(a₁a₂ + b₁b₂)| / (√(a₁² + b₁²) √(a₂² + b₂²)):

a₁a₂ + b₁b₂ = (3)(5) + (4)(-12) = 15 - 48 = -33

√(a₁² + b₁²) = √(3² + 4²) = √(9 + 16) = √25 = 5

√(a₂² + b₂²) = √(5² + (-12)²) = √(25 + 144) = √169 = 13

cos θ = |-33| / (5 * 13) = 33 / 65

The acute angle is θ = cos^-1(33/65).

3. Bisectors of the Angle Between Two Lines

When two lines intersect, they form two pairs of vertically opposite angles. The lines that bisect these angles are called the angle bisectors. There are two angle bisectors, and they are always perpendicular to each other.

Let the two lines be:

Line 1: a₁x + b₁y + c₁ = 0

Line 2: a₂x + b₂y + c₂ = 0

A point (x, y) lies on an angle bisector if its perpendicular distance from both lines is equal (or equal in magnitude).

The distance from (x, y) to Line 1 is |a₁x + b₁y + c₁| / √(a₁² + b₁²).

The distance from (x, y) to Line 2 is |a₂x + b₂y + c₂| / √(a₂² + b₂²).

Equating these distances gives the equations of the angle bisectors:

(a₁x + b₁y + c₁) / √(a₁² + b₁²) = ± (a₂x + b₂y + c₂) / √(a₂² + b₂²)

Determining the Bisector of the Acute/Obtuse Angle:

To determine which sign corresponds to the acute or obtuse angle bisector, we can use the signs of the constant terms and the expression a₁a₂ + b₁b₂.

First, make the constant terms c₁ and c₂ positive. If they are negative, multiply the entire equation by -1.

Let the equations be a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 with c₁ > 0 and c₂ > 0.

1. Case 1: If a₁a₂ + b₁b₂ > 0:
   • The bisector of the obtuse angle is given by the '+' sign: (a₁x + b₁y + c₁) / √(a₁² + b₁²) = + (a₂x + b₂y + c₂) / √(a₂² + b₂²)
   • The bisector of the acute angle is given by the '-' sign: (a₁x + b₁y + c₁) / √(a₁² + b₁²) = - (a₂x + b₂y + c₂) / √(a₂² + b₂²)
2. Case 2: If a₁a₂ + b₁b₂ < 0:
   • The bisector of the obtuse angle is given by the '-' sign: (a₁x + b₁y + c₁) / √(a₁² + b₁²) = - (a₂x + b₂y + c₂) / √(a₂² + b₂²)
   • The bisector of the acute angle is given by the '+' sign: (a₁x + b₁y + c₁) / √(a₁² + b₁²) = + (a₂x + b₂y + c₂) / √(a₂² + b₂²)

Determining the Bisector of the Angle Containing the Origin:

To find the bisector of the angle that contains the origin (0,0), substitute (0,0) into the expressions (a₁x + b₁y + c₁) / √(a₁² + b₁²) and (a₂x + b₂y + c₂) / √(a₂² + b₂²).

Let the equations of the bisectors be:

B₁: (a₁x + b₁y + c₁) / √(a₁² + b₁²) + (a₂x + b₂y + c₂) / √(a₂² + b₂²) = 0

B₂: (a₁x + b₁y + c₁) / √(a₁² + b₁²) - (a₂x + b₂y + c₂) / √(a₂² + b₂²) = 0

Substitute (0,0) into the left side of B₁. If the result is positive, then B₁ is the bisector of the angle containing the origin. If the result is negative, then B₂ is the bisector of the angle containing the origin.

Example:

Find the equations of the bisectors of the angle between the lines 3x - 4y + 7 = 0 and 12x + 5y - 13 = 0.

Line 1: 3x - 4y + 7 = 0. a₁=3, b₁=-4, c₁=7. √(a₁² + b₁²) = √(9 + 16) = 5.

Line 2: 12x + 5y - 13 = 0. a₂=12, b₂=5, c₂=-13. √(a₂² + b₂²) = √(144 + 25) = 13.

The equations of the bisectors are:

(3x - 4y + 7) / 5 = ± (12x + 5y - 13) / 13

Bisector 1 (using + sign):

13(3x - 4y + 7) = 5(12x + 5y - 13)

39x - 52y + 91 = 60x + 25y - 65

(60 - 39)x + (25 + 52)y + (-65 - 91) = 0

21x + 77y - 156 = 0

Bisector 2 (using - sign):

13(3x - 4y + 7) = -5(12x + 5y - 13)

39x - 52y + 91 = -60x - 25y + 65

(39 + 60)x + (-52 + 25)y + (91 - 65) = 0

99x - 27y + 26 = 0

Example 2:

Find the bisector of the angle between x + y - 1 = 0 and x - y + 2 = 0 that contains the origin.

Line 1: x + y - 1 = 0. a₁=1, b₁=1, c₁=-1. √(a₁² + b₁²) = √2.

Line 2: x - y + 2 = 0. a₂=1, b₂=-1, c₂=2. √(a₂² + b₂²) = √2.

The equations of the bisectors are:

(x + y - 1) / √2 = ± (x - y + 2) / √2

x + y - 1 = ± (x - y + 2)

Bisector 1 (+ sign):

x + y - 1 = x - y + 2

2y - 3 = 0

Bisector 2 (- sign):

x + y - 1 = -(x - y + 2)

x + y - 1 = -x + y - 2

2x + 1 = 0

Now, let's check which bisector contains the origin. We need to make constant terms positive first. The equations are already in a form where we can check the constant terms' signs.

For Line 1: x + y - 1 = 0. The constant term is negative. For Line 2: x - y + 2 = 0. The constant term is positive.

Let's rewrite the bisector equations by ensuring constant terms are positive first, or use the direct evaluation method.

Method: Evaluate c₁/√(a₁² + b₁²) ± c₂/√(a₂² + b₂²).

Here, c₁ = -1 and c₂ = 2. Denominators are both √2.

Bisector 1 (using + sign in the formula): (x + y - 1)/√2 + (x - y + 2)/√2 = 0 => 2x + 1 = 0

Bisector 2 (using - sign in the formula): (x + y - 1)/√2 - (x - y + 2)/√2 = 0 => 2y - 3 = 0

Let's test the origin (0,0) in the original line equations to see which angle contains it.

Line 1: 0 + 0 - 1 = -1 (Negative)

Line 2: 0 - 0 + 2 = 2 (Positive)

Since the signs are different, the bisector of the angle containing the origin is given by the equation where the signs of the two expressions are different. This corresponds to the '-' sign in the general formula: (a₁x + b₁y + c₁) / √(a₁² + b₁²) - (a₂x + b₂y + c₂) / √(a₂² + b₂²) = 0.

This leads to 2y - 3 = 0.

Therefore, the bisector of the angle containing the origin is 2y - 3 = 0.