Two distinct lines intersect at the most at one point. To discover the intersection of 2 lines we require the general form of the two equations, which is written as (eginarrayla_1x + b_1y + c_1 = 0, ext and a_2x + b_2y + c_2 = 0endarray). The lines will certainly intersect just if they room non-parallel lines. Common examples of intersecting currently in real life encompass a pair of scissors, a urgent chair, a road cross, a signboard, etc. In this mini-lesson, we will learn in detail, exactly how to discover the point of intersection of 2 lines.

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1. | Meaning the Intersection of 2 Lines |

2. | Finding Intersection of 2 Lines |

3. | The edge of Intersection |

4. | Conditions for two Lines to be Parallel or Perpendicular |

5. | Properties of Intersecting Lines |

6. | Solved instances on Intersection of 2 Lines |

7. | Practice concerns on Intersection of two Lines |

8. | FAQS on Intersection of 2 Lines |

## Meaning the Intersection of two Lines

When two lines share specifically one common point, lock are called the intersecting lines. The intersecting lines re-superstructure a common point. And, this common allude that exists on every intersecting present is dubbed the allude of intersection. The two non-parallel directly lines which space co-planar will have actually an intersection point. Here, lines A and B intersect at suggest O, i beg your pardon is the point of intersection.

## Finding Intersection of 2 Lines

Let"s take into consideration the complying with case. We are offered two lines, (L_1) and (L_2), and also we are forced to uncover the allude of intersection. Evaluating the point of intersection involves solving two simultaneous direct equations.

Let the equations of the two lines be (written in the general form): (eginarrayla_1x + b_1y + c_1 = 0\a_2x + b_2y + c_2 = 0endarray)

Now, let the point of intersection be (left( x_0,y_0 ight)). Thus,

(eginarrayla_1x_0 + b_1y_0 + c_1 = 0\a_2x_0 + b_2y_0 + c_2 = 0endarray)

This system can be addressed using Cramer’s preeminence to get:

(fracx_0b_1c_2 - b_2c_1 = frac - y_0a_1c_2 - a_2c_1 = frac1a_1b_2 - a_2b_1)

From this relation, us can achieve the suggest of intersection (left( x_0,y_0 ight)) as

(left( x_0,y_0 ight) = left( fracb_1c_2 - b_2c_1a_1b_2 - a_2b_1,fracc_1a_2 - c_2a_1a_1b_2 - a_2b_1 ight))

## The edge of Intersection

To attain the edge of intersection between two lines, take into consideration the figure shown:

The equations the the two lines in slope-intercept form are:

(eginalign&y = left( - fraca_1b_1 ight)x + left( fracc_1b_1 ight) = m_1x + C_1\&y = left( - fraca_2b_2 ight)x + left( fracc_2b_2 ight) = m_2x + C_2endalign)

Note in the figure above that ( heta = heta _2 - heta _1), and also thus

(eginalign& an heta = an left( heta _2 - heta _1 ight) = frac an heta _2 - an heta _11 + an heta _1 an heta _2\&qquadqquadqquadqquad;;= fracm_2 - m_11 + m_1m_2endalign)

Conventionally, we would certainly be interested only in the acute angle in between the 2 lines and thus, we need to have ( an heta ) as a positive quantity.

So in the expression above, if the expression (fracm_2 - m_11 + m_1m_2) turns out to be negative, this would be the tangent the the obtuse angle between the 2 lines; thus, to gain the acute angle in between the two lines, we usage the size of this expression.

Therefore, the acute angle ( heta ) between the 2 lines is

( heta = an ^ - 1left| fracm_2 - m_11 + m_1m_2 ight|)

From this relation, we can conveniently deduce the problems on (m_1) and also (m_2) such the the two lines (L_1) and also (L_2) are parallel or perpendicular.

## Conditions for 2 Lines to be Parallel or Perpendicular

If the lines room parallel, ( heta = 0) and (m_1 = m_2), which is evident since parallel currently must have the exact same slope.

For the two lines to it is in perpendicular lines, θ = π/2 , so that cot θ = 0; this can happen if (1 + m_1m_2 = 0) or (m_1m_2 = - 1).

If the currently (L_1) and also (L_2) space in the general kind ax + by + c = 0, the steep of this line is m = -a/b.

### Condition for 2 Lines to it is in Parallel

Thus, the condition for (L_1) and (L_2) to be parallel is:

(m_1 = m_2, Rightarrow , - fraca_1b_1 = - fraca_2b_2, Rightarrow ,fraca_1b_1 = fraca_2b_2)

### Example

The heat (L_1:x - 2y + 1 = 0) is parallel to the heat (L_2:x - 2y - 3 = 0) due to the fact that the steep of both the currently is m = 1/2

### Condition for two Lines to it is in Perpendicular

The condition for (L_1) and also (L_2) to be perpendicular is:

(eginalign&m_1m_2 = - 1, Rightarrow ,left( - fraca_1b_1 ight)left( - fraca_2b_2 ight) = - 1,\ &qquadqquad;;;; Rightarrow ,,a_1a_2 + b_1b_2 = 0endalign)

### Example

The heat (L_1) : x + y = 1 is perpendicular come the heat (L_2) :x - y = 1 because the steep of (L_1) is ( - 1) while the slope of (L_2) is 1.

## Properties the Intersecting Lines

The intersecting currently (two or more) constantly meet in ~ a solitary point.The intersecting lines deserve to cross each other at any kind of angle. This angle created is always greater 보다 0∘ and less than 180∘. Two intersecting lines kind a pair of vertical angles. The upright angles space opposite angles through a common vertex (which is the suggest of intersection).

Here, ∠a and also ∠c are vertical angles and are equal.

Also, ∠b and ∠d are vertical angles and also equal to every other.

∠a+∠d = right angle =180∘

An acute edge ( heta ) between present (L_1) and also (L_2) v slopes (m_1) and (m_2) is given by

( heta = an ^ - 1left| fracm_2 - m_11 + m_1m_2 ight|)

If the present (L_1) and (L_2) are given in the basic form ax + by + c = 0, the slope of this line is m = -a/b.

The problem for two lines (L_1) and also (L_2) to it is in parallel is:(m_1 = m_2)The problem for 2 lines (L_1) and also (L_2) to it is in perpendicular is:(m_1m_2 = - 1)

### Related short articles on Intersection of 2 Lines

Check the end the write-ups below to know much more about topics pertained to the intersection of 2 lines.

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**Example 1: find the point of intersection and the edge of intersection for the following two lines:**

x - 2y + 3 = 03x - 4y + 5 = 0

**Solution:**

We use Cramer’s ascendancy to discover the point of intersection:

x/(-10 - (-12)) = -y/(5-9) = 1/(-4 - (-6))

⇒ x/2 = y/4 = 1/2

⇒ x = 1, y = 2Now, the slopes of the 2 lines are:

(m_1 = frac12,,,,m_2 = frac34)

If ( heta ) is the acute edge of intersection in between the 2 lines, we have:

(eginalign& an heta = left| fracm_2 - m_11 + m_1m_2 ight| = left| fracfrac34 - frac121 + frac38 ight| = frac211endalign)

θ = tan−1(2/11) ≈ 10.3∘

**∴** Point of intersection is (1,2).The angle of intersection is θ = tan−1(2/11)

**Example 2: find the equation that a line perpendicular come the heat x - 2y + 3 = 0 and passing with the allude (1, -2).**

**Solution:**

Given line x - 2y + 3 = 0 have the right to be created as

y = (1/2)x + 3/2

Slope that the heat 1 is (m_1)= 1/2

Therefore, slope of the heat perpendicular to heat ((1)) is

(m_2 = -frac1m_1 ) = -2

Equation the a heat perpendicular to the line x - 2y + 3 = 0 and passing with the allude (1, -2) is