Two lines intersection calculator Print intersection of 2 planes calculator
Line 1 Equation
x + y + = 0
2 points
(x1,y1) ( ) (x2,y2) ( )
Slope, Point
(xp,yp) ( ) Slope = Angle =
Line 2 Equation
x + y + = 0
2 points
(x1,y1) ( ) (x2,y2) ( )
Slope, Point
(xp,yp) ( ) Slope = Angle =
Angle between lines
Intersection point
x =
y =
Angle
Bisector
lines
Equations
Slopes
Angles
Degree Radian    
line geometry Inclined lines
Two lines intersection Print two lines intersection summary
Lines given by the equations:
y = m1x + a
y = m2x + b
Where in vector notation:
A = m1i - j
B = m2i − j
The intersection point is determined by solving the values of x and y from the two lines equations by Cramer's rule or by direct substitution:
If     m2 − m1 = 0     then both lines are parallel.
The angle between two lines in the range   0 < θ < π/2   is:
Angle between two lines

The angle between the two lines can be calculated by vector dot product method:       A · B = |A| |B|cos θ
Note: the ± sign stands for the two possible angles between two lines that are complementary to  180 degrees.
If the two lines are given by the equations:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
The intersection point is determined by solving the values of x and y from the two lines equations:
If     a1b2 − a2b1 = 0     then both lines are parallel.
If both lines are each given by two points, first line points:     (x1 , y1) , (x2 , y2) and the second line is given by two points:
(x3 , y3) , (x4 , y4)
The intersection point (x , y) is found by the equations:
Example: find the intersection point and the angle between the lines:
x − 2y + 3 = 0
3x + 4y + 1 = 0
Solving the lines equations for x and y by Cramer's rule.
And the intersection point is at (− 1.4 , 0.8)
The angle between the lines is:
If the two solutions are added then:   63.43 + 116.57 = 180   as we expected.
Two lines angle bisector Print two lines angle bisector summary
Two lines bisector angle
Lines given by the equations:
y = m1x + a
y = m2x + b
The angle of the lines angle bisector from the x axis is:
The equation of the angle bisector line is:
First line:
A second angle bisector exists at a right angle from the first line.
Second line:
Where:
mb can be expressed by the slopes m1 and m2 of the lines as:
The ± sign stands for the two angle bisectors possible between two lines (complimantery to 180 degree).
Angle bisector line equation expressed by m1 and m2 is:
y − y0 = mb(x − x0)
If the lines are given by the equations:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
The distance (d) of any common point (x , y) on the angle bisector between two lines are the same (two similar triangles).
distance of point (x , y) from line (1)
distance of point (x , y) from line (2)
And the equation of the lines angle bisector is:
This equation can be written as:
(a1 − φ a2)x + (b1 − φ b2)y + (c1 − φ c2) = 0
Where:
Note: The Plus and minus sign is used to find the two possible angle bisectors lines equation which are  90  degrees apart.
Two lines angle bisector example Print two lines angle bisector example
Example: given two lines:
3x − 4y + 2 = 0
5x + 12y + 1 = 0
find the angle between the lines and the equation of the angle bisector between the two lines.
The angle between the lines is found by vector dot product method.
We can write the lines general direction by vector notation as:
L1 = a1i + b1j       and       L2 = a2i + b2j
The dot product of this two vectors are related to the angle by the formula:       L1 · L2 = |L1||L2|cos θ
Where:
L1 · L2 = a1a2 + b1b2
Then the two possible angles are:
In order to find the angle bisector line equation we use the distance equations:
And the two lines angle bisector lines equation are:
13(3x − 4y +2) = 5(5x + 12y + 1)
14x − 112y + 21 = 0
And the second line:
13(3x − 4y + 2) = 5(5x + 12y + 1)
64x + 8y + 31 = 0
Example 3 - intersection of 2 lines each defined by 2 points Print example of intersection of 2 lines defines by 4 points
Find the intersection point of two lines each of them defined by two pair of coordinates, first line by
(x1 y1) (x2 y2) and second line by (x3 y3) and (x4 y4).
The equations of two arbitrary lines are:    y = m1 x + a  and  y = m2 x + b
This equation can be easily solved by comparing the y values of both equations: m1 x + a = m2 x + b
And the solution is: x=(a-b)/(m_2-m_1 ) and y=(m_1 (a-b))/(m_2-m_1 )+a
Line slope
from the schetch the slope  m1  is:   m_1=tan⁡θ=(y_2-y_1)/(x_2-x_1 )
the same solution for slope  m2  is:   m_1=tan⁡θ=(y_2-y_1)/(x_2-x_1 )
The slope m1 can be written as a function of x and y as:
m_1=(y-y_1)/(x-x_1 ) or m_1 (x-x_1 )=y-y_1
And the line equation will be: y=m_1 x+(y_1-m_1 x_1 )
We can see that the free item that is  a  in the line equation is equal to  a = y1 − m1x1  and for the second line  b = y3 − m2x3
Now that we have found all the values of m1 m2 a  and  b,  we can substitute those values to the  x  and  y  coordinate of the intersection point.
x=(a-b)/(m_2-m_1 )=(y_1-(y_2-y_1)/(x_2-x_1 ) x_1-y_3+(y_4-y_3)/(x_4-x_3 ) x_3)/((y_4-y_3)/(x_4-x_3 )-(y_2-y_1)/(x_2-x_1 ))
=(((x_2-x_1 )(x_4-x_3 )(y_1-y_3 )-(x_4-x_3 )(x_1 y_2-x_1 y_1 )+(x_2-x_1 )(x_3 y_4-x_3 y_3 ))/(x_2-x_1 )(x_4-x_3 ) )/(((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ))/(x_2-x_1 )(x_4-x_3 ) )
=((x_2-x_1 )(x_4-x_3 )(y_1-y_3 )-(x_4-x_3 )(x_1 y_2-x_1 y_1 )+(x_2-x_1 )(x_3 y_4-x_3 y_3 ))/((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ) )
=(x_2 x_4 y_1-x_2 x_3 y_1-x_2 x_4 y_3+x_1 x_4 y_3-x_4 x_1 y_2+x_3 x_1 y_2+x_2 x_3 y_4-x_1 x_3 y_4)/((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ) )
=(x_2 y_1 (x_4-x_3 )-x_4 y_3 (x_2-x_1 )-x_1 y_2 (x_4-x_3 )-x_3 y_4 (x_2-x_1 ))/((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ) )
x=((x_2 y_1-x_1 y_2 )(x_4-x_3 )-(x_4 y_3-x_3 y_4 )(x_2-x_1 ))/((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ) )
And the same process accompanied with some basic algebra steps we get the y coordinate:
y=(m_1 (a-b))/(m_2-m_1 )+a=((y_2-y_1)/(x_2-x_1 )  (y_1-m_1 x_1-y_3+m_2 x_3 ))/((y_4-y_3)/(x_4-x_3 )-(y_2-y_1)/(x_2-x_1 ))+y_1-m_1 x_1
=(m_2 (y_1-m_1 x_1 )-m_1 (y_3-m_2 x_3 ))/(m_2-m_1 )=(m_2 y_1-m_1 m_2 x_1-m_1 y_3+m_1 m_2 x_3)/(m_2-m_1 )
=(m_1 m_2 (x_3-x_1 )+m_2 y_1-m_1 y_3)/(m_2-m_1 )
Now substitute the values of m1 and m2 to get:
=((y_2-y_1)/(x_2-x_1 )  (y_4-y_3)/(x_4-x_3 ) (x_3-x_1 )+(y_4-y_3)/(x_4-x_3 ) y_1-(y_2-y_1)/(x_2-x_1 ) y_3)/((y_4-y_3)/(x_4-x_3 )-(y_2-y_1)/(x_2-x_1 ))
=(((y_2-y_1 ))/((x_2-x_1 ) )  ((y_4-y_3 ))/((x_4-x_3 ) ) (x_3-x_1 )+(x_2-x_1 )(y_4-y_3 )/(x_2-x_1 )(x_4-x_3 )  y_1-(x_4-x_3 )(y_2-y_1 )/(x_4-x_3 )(x_2-x_1 )  y_3)/(((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ))/(x_2-x_1 )(x_4-x_3 ) )
=((y_2-y_1 )(y_4-y_3 )(x_3-x_1 )+(x_2 y_1-x_1 y_1 )(y_4-y_3 )-(x_4 y_3-x_3 y_3 )(y_2-y_1 ))/((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ) )
=(x_3 y_2 y_4-x_3 y_1 y_4-x_1 y_2 y_4+x_1 y_2 y_3+x_2 y_1 y_4-x_2 y_1 y_3-x_4 y_2 y_3+x_4 y_1 y_3)/((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ) )
=(y_4 (x_2 y_1-x_1 y_2 )-y_3 (x_2 y_1-x_1 y_2 )-y_2 (x_4 y_3+x_3 y_4 )+y_1 (x_4 y_3-x_3 y_4 ))/((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ) )
y=((x_2 y_1-x_1 y_2 )(y_4-y_3 )+(x_4 y_3+x_3 y_4 )(y_2-y_1 ))/((x_2-x_1 )(y_4-y_3 )-(x_4-x_3 )(y_2-y_1 ) )