When two lines come together at a point, it's called their intersection. This happens when the lines are not parallel to each other. The point where two lines meet is called their point of intersection. In other words, when two lines share a common point, they are called intersecting lines.

We can find where two lines meet by solving their equations or by drawing them on a graph and seeing where they cross. Real-life examples of intersecting lines include two roads intersecting in a traffic signal, intersecting lines in the goal box of a soccer field, beams intersecting at various angles in a bridge, intersecting streets laid out on a city map, etc. In this lesson, we'll go through how to find where two lines intersect.

To find where two lines meet, we follow an algebraic process. Let's take two lines, \(L_1\) and \(L_2\), as an example. Our goal is to figure out where they intersect. This involves solving the equations representing \(L_1\) and \(L_2\) simultaneously.

First, we write the equations of the lines in a standard form. Let's say the equations look like this:

\(a_1x + b_1y + c_1 = 0\)

\(a_2x + b_2y + c_2 = 0\)

Now, let's assume that the point where they intersect is \((x_0, y_0)\). The point of intersection lies on both the intersecting lines. This means that the point of intersection \((x_0, y_0)\) satisfies the equations of both lines. So, we have:

\(a_1x_0 + b_1y_0 + c_1 = 0\)

\(a_2x_0 + b_2y_0 + c_2 = 0\)

By using a rule called Cross Multiplication, we can solve this system of equations. It helps us find \(x_0\) and \(y_0\). The formulas look like this:

`x_0 = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}`

`y_0 = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}`

To find the angle where two lines meet, we consider a diagram. In this diagram, we have two lines intersecting at a point, and we want to calculate the angle between them.

First, we express the equations of these lines in slope-intercept form. These equations help us understand how the lines are positioned relative to the `x`-axis.

In the diagram, the angle \( \theta \) between the lines is \( \theta = \theta_2 - \theta_1 \), where \( \theta_1 \) and \( \theta_2 \) are the angles each line makes with the `x`-axis.

We use trigonometric functions to calculate the tangent of \( \theta \), which is given by `\tan \theta = \tan (\theta_2 - \theta_1) = \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_1 \tan \theta_2}`.

Since \( \tan \theta_1 = m_1 \) and \( \tan \theta_2 = m_2 \), we can write

`\frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_1 \tan \theta_2} = \frac{m_2 - m_1}{1 + m_1 m_2}`

Conventionally, we're interested in the acute angle between the lines. So, if the expression `\frac{m_2 - m_1}{1 + m_1 m_2}` is negative, it means we're dealing with the obtuse angle. To ensure we get the acute angle, we take the magnitude of this expression.

Thus, the acute angle \( \theta \) between the lines is given by `\theta = \tan^{-1} | \frac{m_2 - m_1}{1 + m_1 m_2}|`.

This formula helps us determine whether the lines are parallel or perpendicular based on the slopes \( m_1 \) and \( m_2 \).

- Intersecting lines always meet at a single point.
- These lines can cross each other at any angle between `0°` and `180°`.
- When two lines intersect, they form a pair of vertical angles, which are opposite angles sharing the point of intersection.
- An acute angle \( \theta \) between two lines \( L_1 \) and \( L_2 \) with slopes \( m_1 \) and \( m_2 \) is given by `\theta = \tan^{-1} | \frac{m_2 - m_1}{1 + m_1 m_2}|`.
- Two lines \( L_1 \) and \( L_2 \) are parallel if their slopes are equal: \( m_1 = m_2 \).
- Two lines \( L_1 \) and \( L_2 \) are perpendicular if the product of their slopes equals `-1`: \( m_1 m_2 = -1 \).

To determine if two lines are parallel or perpendicular, we consider the angle between them.

When lines are parallel, the angle \( \theta \) between them is `0`, meaning they don't slant away from each other. This occurs when their slopes \( m_1 \) and \( m_2 \) are equal, as parallel lines have the same slope.

For lines to be perpendicular, the angle \( \theta \) between them is a right angle, or `\frac{\pi}{2}` radians. This happens when \( 1 + m_1 m_2 = 0 \) or \( m_1 m_2 = -1 \).

Another way to express these conditions is by using the slope of a line formula, where the slope \( m \) of a line \( ax + by + c = 0 \) is `-a/b`. Applying this to two lines \( L_1 \) and \( L_2 \), their slopes \( m_1 \) and \( m_2 \) help determine if they're parallel or perpendicular.

To determine if two lines \( L_1 \) and \( L_2 \) are parallel, we check if their slopes are equal. This means that the coefficient of \( x \) in the equation of \( L_1 \) divided by the coefficient of \( y \) is equal to the same ratio in the equation of \( L_2 \).

For example, if the slope of \( L_1 \) is `\frac{1}{2}`, and the slope of \( L_2 \) is also `\frac{1}{2}`, then the lines are parallel.

An instance of this is when the line \( L_1 \) with equation \( x - 2y + 1 = 0 \) is parallel to the line \( L_2 \) with equation \( x - 2y - 3 = 0 \). This happens because both lines have a slope of `\frac{1}{2}`.

To determine if two lines \( L_1 \) and \( L_2 \) are perpendicular, we check if the product of their slopes is equal to `-1`. This means that the negative reciprocal of the slope of one line is equal to the slope of the other.

For example, if the slope of \( L_1 \) is `1/2` and the slope of \( L_2 \) is `-2`, then the lines are perpendicular.

An example of this is when the line \( L_1 \) with equation \( -x + 2y = 1 \) having slope `1/2` is perpendicular to the line \( L_2 \) with equation \( -2x - y = 1 \) having slope `-2`. This happens because `1 + (1/2)(-2) = 0`, satisfying the condition for perpendicularity \( 1 + m_1 m_2 = 0 \)

To find the point of intersection of two lines, we can utilize the method of solving a system of equations. There are primarily two methods for solving such systems: the substitution method and the elimination method.

**Substitution Method:**

In the substitution method, we solve one of the equations for one variable and then substitute this expression into the other equation. This allows us to solve for the remaining variable. Once we have found the value of one variable, we can substitute it back into either of the original equations to find the corresponding value of the other variable.

**Example:**

**Consider the system of equations:**

**\(2x + y = 5\)**

**\(3x - 2y = 8\)**

**Solution: **

Using the substitution method, we can solve the first equation for \(y\):

\(y = 5 - 2x\)

Now, we substitute this expression for \(y\) into the second equation:

\(3x - 2(5 - 2x) = 8\)

\(3x - 10 + 4x = 8\)

\(7x - 10 = 8\)

\(7x = 18\)

`x = \frac{18}{7}`

Substituting `x = \frac{18}{7}` into the first equation:

`2 \left( \frac{18}{7} \right) + y = 5`

`y = 5 - \frac{36}{7}`

`y = -\frac{1}{7}`

So, the point of intersection is `\left( \frac{18}{7}, -\frac{1}{7} \right)`.

**Elimination Method:**

In the elimination method, we manipulate the equations to eliminate one of the variables by adding or subtracting the equations. This results in a single equation with one variable, which we can then solve to find its value. Once we have the value of one variable, we can substitute it back into one of the original equations to find the corresponding value of the other variable.

**Example:**

**Consider the system of equations:**

**\(3x + 2y = 10\)**

**\(2x - y = 4\)**

**Solution:**

To eliminate \(y\), we can multiply the second equation by `2` and add it to the first equation:

\(3x + 2y + 4x - 2y = 10 + 8\)

\(7x = 18\)

`x = \frac{18}{7}`

Substituting `x = \frac{18}{7}` into the second equation:

`2 \left( \frac{18}{7} \right) - y = 4`

`y = 2 \left( \frac{18}{7} \right) - 4`

`y = \frac{8}{7}`

So, the point of intersection is `\left( \frac{18}{7}, \frac{8}{7} \right)`.

**Example `1`: Find the point of intersection for the lines \(3x + 2y - 5 = 0\) and \(2x - y + 3 = 0\).**

**Solution: **

Using the method described, let's find the point of intersection for the lines \(3x + 2y - 5 = 0\) and \(2x - y + 3 = 0\).

First, we identify the coefficients \(a_1\), \(b_1\), \(c_1\), \(a_2\), \(b_2\), and \(c_2\):

\(a_1 = 3, \quad b_1 = 2, \quad c_1 = -5\)

\(a_2 = 2, \quad b_2 = -1, \quad c_2 = 3\)

Now, we apply the formulas for \(x_0\) and \(y_0\):

`x_0 = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}`

`y_0 = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}`

Substituting the values:

`x_0 = \frac{(2)(3) - (-1)(-5)}{(3)(-1) - (2)(2)}`

`y_0 = \frac{(-5)(2) - (3)(3)}{(3)(-1) - (2)(2)}`

Solving these equations:

`x_0 = \frac{6 - 5}{-3 - 4} = -\frac{1}{7}`

`y_0 = \frac{-10 - 9}{-3 - 4} = \frac{-19}{-7} = \frac{19}{7}`

So, the point of intersection is `(-\frac{1}{7}, \frac{19}{7})`.

**Example `2`: Determine if the lines \(4x - 2y + 7 = 0\) and \(2x - y + 3 = 0\) are parallel, perpendicular, or neither.**

**Solution: **

The slopes of the lines are:

`m_1 = \frac{4}{2} = 2`

`m_2 = \frac{2}{1} = 2`

Since the slopes are equal, the lines are parallel.

**Example `3`: Find the acute angle between the lines \(2x + 3y - 4 = 0\) and \(3x + 2y + 5 = 0\).**

**Solution: **

The slopes of the lines are:

`m_1 = \frac{-2}{3}`

`m_2 = \frac{-3}{2}`

Using the formula for the acute angle between two lines:

`\theta = \tan^{-1} | \frac{m_2 - m_1}{1 + m_1 m_2}|`

`\theta = \tan^{-1} | \frac{(\frac{-3}{2}) - (\frac{-2}{3})}{1 + (\frac{-2}{3}) (\frac{-3}{2})}|`

`\theta = \tan^{-1} | \frac{-\frac{5}{6}}{1 + 1}|`

`\theta = \tan^{-1} | \frac{-\frac{5}{6}}{2}|`

`\theta = \tan^{-1} |-\frac{5}{12}|`

`\theta = \tan^{-1} ( \frac{5}{12})`

Calculating this, we find \( \theta \approx 22.62^\circ \).

So, the acute angle between the lines is approximately \( 22.62^\circ \).

**Example `4`: Determine if the lines \(5x + 2y - 3 = 0\) and \(2x - 5y + 1 = 0\) are parallel, perpendicular, or neither.**

**Solution: **

To determine the relationship between the lines, we compare their slopes.

The slope of the first line is `\frac{-5}{2}` and the slope of the second line is `\frac{2}{5}`.

The product of their slopes is `-1`. Hence the lines are perpendicular.

**Example `5`: Find the point of intersection for the lines \(x + 2y - 5 = 0\) and \(3x - y + 4 = 0\) using the elimination method.**

**Solution: **

To find the point of intersection for the lines \(x + 2y - 5 = 0\) and \(3x - y + 4 = 0\) using the elimination method, we can manipulate the equations to eliminate one of the variables.

First, let's rewrite the equations in standard form:

\(x + 2y = 5\)

\(3x - y = -4\)

Now, to eliminate \(y\), we can multiply the first equation by `1` and the second equation by `2` to make the coefficients of \(y\) the same:

\(1(x + 2y) = 1(5)\)

\(2(3x - y) = 2(-4)\)

This gives us:

\(x + 2y = 5\)

\(6x - 2y = -8\)

Adding these equations together, we eliminate \(y\):

\(x + 2y + 6x - 2y = 5 - 8\)

\(7x = -3\)

Now, we can solve for \(x\):

`x = \frac{-3}{7}`

Substituting `x = \frac{-3}{7}` into either of the original equations, let's use the first one:

\(x + 2y = 5\)

`\frac{-3}{7} + 2y = 5`

`2y = \frac{38}{7}`

`y = \frac{38}{14}`

`y = \frac{19}{7}`

So, the point of intersection is `(\frac{-3}{7}, \frac{19}{7})`.

**Q`1`. Determine if the lines \(3x + 4y - 6 = 0\) and \(2x - 3y + 5 = 0\) are parallel, perpendicular, or neither.**

- Parallel
- Perpendicular
- Neither

**Answer:** c

**Q`2`. Find the acute angle between the lines \(6x + 3y - 9 = 0\) and \(4x - 2y + 6 = 0\).**

- \(45.72^\circ\)
- \(53.13^\circ\)
- \(62.5^\circ\)
- \(75.45^\circ\)

**Answer:** b

**Q`3`. Find the point of intersection for the lines \(x - 2y + 3 = 0\) and \(2x - 3y + 6 = 0\).**

- \((-3, 0)\)
- \((1, -2)\)
- \((2, 1)\)
- \((3, -4)\)

**Answer:** a

**Q`4`. Find the equation of the line passing through the point \((2, 3)\) and perpendicular to the line \(4x - 2y + 8 = 0\).**

- \(4x - 2y - 5 = 0\)
- \(2x - 4y - 5 = 0\)
- \(4x + 2y - 7 = 0\)
- \(x + 2y - 8 = 0\)

**Answer:** d

**Q`5`. Find the coordinates of the point where the lines \(3x + 5y - 7 = 0\) and \(5x - 3y + 11 = 0\) intersect.**

- \((-1, -2)\)
- \((1, -2)\)
- \((-1, 2)\)
- \((1, 2)\)

**Answer:** c

**Q`1`. What is the intersection point of two lines?**

**Answer:** The intersection point of two lines is the point where both lines meet or cross each other. It is the common solution to the equations representing the two lines.

**Q`2`. How do you determine if two lines are parallel, perpendicular, or neither?**

**Answer:** Two lines are parallel if their slopes are equal. They are perpendicular if the product of their slopes is `-1`. If neither condition is met, the lines are neither parallel nor perpendicular.

**Q`3`. How can we find the equation of a line passing through a given point and perpendicular to another line?**

**Answer:** To find the equation of a line passing through a given point and perpendicular to another line, we first determine the slope of the given line. Then, we find the negative reciprocal of this slope, which becomes the slope of the perpendicular line. Finally, we use the point-slope form to find the equation of the perpendicular line.

**Q`4`. What is the acute angle between two intersecting lines?**

**Answer:** The acute angle between two intersecting lines is the smallest angle formed by the lines when they meet. It can be calculated using trigonometric formulas based on the slopes of the lines.

**Q`5`. How do you find the point of intersection of two lines?**

**Answer:** One of the ways to find the point of intersection of two lines is to solve the system of equations representing the two lines. The solution gives the coordinates of the point where the lines intersect.