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Triangle defined by 3 lines

Lines form:      ax + by + c = 0
Line 1: x + y + = 0
Line 2: x + y + = 0
Line 3: x + y + = 0
Intersection Coordinates (x , y): Line 1-2
Line 1-3
Line 2-3
Angle Between Lines: Line 1-2
Line 1-3
Line 2-3
Lines Length: Line 1
Line 2
Line 3
Angle to x axis: Line 1
Line 2
Line 3
Triangle Area:
Triangle Circumference:
Intersection point of the medians:
Intersection point of the altitudes:
Intersection point of the perpendicular bisectors:
Intersection point of the angle bisectors (incircle center):
Incircle equation:
Circumcircle equation:
       Degree Radian
Line geometry
Triangle defined by 3 points
Triangle overview
Point inside a triangle
Triangle ex1

Triangle defined by three lines

Print triangle defined by three lines

Triangle lines equations form:
Area (A)

Intersection points of the triangle sides:

Angles between the triangle sides:

Intersection point of the medians (x , y)
(centroid - also known as the center of gravity)
We use the sides intersection points that had been found before.

Triangle given by three lines of the form:

Intersection points of the triangle sides:

Lines 1,2
Lines 2,3
Lines 3,1
x_1=(b_1 c_2-b_2 c_1)/(a_1 b_2-a_2 b_1 ) y_1=(a_2 c_1-a_1 c_2)/(a_1 b_2-a_2 b_1 )

Example 1

Print triangle defined by three lines Example1
Q
Find the vertices, and the center of the circumcircle of the triangle formed by the equations:
1)  x + 2y = 1       2)   2x − 2y = 5       3)   x + 4y = 2
S
Given triangle

The vertices are the intersection point of all three pairs of the triangle sides:

Intersection points of the sides according to Cremer’s rule or by substitution method.

line 1) and 2)
x=|■(1&2@5&-2)|/|■(1&2@2&-2)| =(-2-10)/(-2-4)=2
y=|■(1&1@2&5)|/|■(1&2@2&-2)| =(5-2)/(-2-4)=-0.5
line 1) and 3)
x=|■(1&2@2&4)|/|■(1&2@1&4)| =(4-4)/(4-2)=0
y=|■(1&1@1&2)|/|■(1&2@1&4)| =(2-1)/(4-2)=0.5
line 2) and 3)
x=|■(5&-2@2&4)|/|■(2&-2@1&4)| =(20+4)/(8+2)=2.4
y=|■(2&5@1&2)|/|■(2&-2@1&-4)| =(4-5)/(8+2)=-0.1

And the vertices are at the points:     (2 , −0.5)     (0 , 0.5)     (2.4 , −0.1)

the length of the triangle sides are the distances between the vertices:

Length of Side 1) 2):
d_1=√((2-0)^2+(-0.5-0.5)^2 )=√5=2.24
Length of Side 1) 3):
d_2=√((2-2.4)^2+(-0.5+0.1)^2 )=√0.32=0.57
Length of Side 2) 3):
d_3=√((0-2.4)^2+(0.5+0.1)^2 )=√6.12=2.47

The circumcircle center is located at the point of the intersection of the perpendicular bisectors of the triangle ribs.

The midpoint of the ribs can be found by the equation

x_m1=(x_a+x_b)/2 y_m1=(y_a+y_b)/2

Substituting values of the end points of line 1) we get the center point:

x_m1=(0+2.4)/2=1.2 y_m1=(0.5-0.1)/2=0.2

The slope of the perpendicular line to side 1) is:       mp1 = −1/m1 = −1/−0.5 = 2

The equation of the perpendicular line that is passing through the midpoint (1 , 0) and has slope of   mp1 = 2 is:

y-y_m1=m_p1 (x-x_m1)

y − 0 = 2 (x − 1)

y = 2x − 2

We will perform the same process on line 2 to get the mid point of line 2:

x_m1=(2.4+2)/2=2.2 y_m1=(-0.1-0.5)/2=-0.3

The slope of the perpendicular line to side 2) is:       mp2 = −1/m2 = −1/1 = −1

And the second perpendicular line equation is:       y = −x + 1.9

The intersection of the 2 lines   y = 2x − 2   and   y = −x + 1.9   will evaluate the coordinate of the circumcircle center:

After solving for x and y we get the coordinates of the circumcenter:  (1.3 , 0.6)

The radius of the circumcircle is the distance between the point of the triangle sides intersection to the circumcenter  (2 , -0.5)  and  (1.3 , 0.6)

R=√((1-1.3)^2+(0-0.6)^2 )=√0.45=0.67

The equation of the circumcircle is:
(x-1.3)^2+(y-0.6)^2=〖1.3〗^2