Question

find incentre of triangle x+1=0,3x-4y=5,5x+2y=27

Answer 1

If we want to find the Incircle radius "r", then we can find vertices A (x1,y1), B(x2, y2), C(x3, y3), lengths of sides BC = a, CA = b, AB = c, and then area  Δ of the triangle ABC.  Then 
                 r = Δ / s      where s = (a+b+c)/2

If we want to find the Incenter I (xin, yin) then
       x_in = [a x1 + b x2 + c x3] / (a + b + c)
       y_in = [a y1 + b y2 + c y3) / (a + b + c)

The sides of the triangle are given by the equations of straight lines:
  AB:  x + 1 = 0         =>  x = -1
  BC:  3 x - 4y = 5     =>  y = 3/4 x - 5/4
  CA :  5x + 2y = 27  =>  y = -5/2  x - 27/2

Solve the above three equations in pairs to get:
   Intersection of AB and CA :  A (x1, y1) = (-1, 16)
   Intersection of  AB and BC :   B (x2 , y2) = (-1, -2)
   Intersection of  BC and CA :  C (x3 , y3) = (59/13, 28/13)

lengths of sides
$BC=a=\sqrt{(59/13+1)^2+(28/13+2)^2}=\frac{90}{13}\\\\CA=b=\sqrt{(59/12+1)^2+(16-28/13)^2}=\frac{36\sqrt{29}}{13}\\\\AB=c=\sqrt{(-1+1)^2+(16+2)^2}=18$

s=(a+b+c)/2 = [18 + 2√29]*9/13 = 19.918....

x_in = (a * x1 + b * x2 + c * x3)  / (a+b+c) = 3/2
y_in  = (a * y1 + b * y2 + c * y3) / (a+b+c) = 3

Answer 2

Answer:

If we want to find the Incircle radius "r", then we can find vertices A (x1,y1), B(x2, y2), C(x3, y3), lengths of sides BC = a, CA = b, AB = c, and then area Δ of the triangle ABC. Then

r = Δ / s where s = (a+b+c)/2

If we want to find the Incenter I (xin, yin) then

x_in = [a x1 + b x2 + c x3] / (a + b + c)

y_in = [a y1 + b y2 + c y3) / (a + b + c)

The sides of the triangle are given by the equations of straight lines:

AB: x + 1 = 0 => x = -1

BC: 3 x - 4y = 5 => y = 3/4 x - 5/4

CA : 5x + 2y = 27 => y = -5/2 x - 27/2

Solve the above three equations in pairs to get:

Intersection of AB and CA : A (x1, y1) = (-1, 16)

Intersection of AB and BC : B (x2 , y2) = (-1, -2)

Intersection of BC and CA : C (x3 , y3) = (59/13, 28/13)

lengths of sides

\begin{gathered}BC=a=\sqrt{(59/13+1)^2+(28/13+2)^2}=\frac{90}{13}\\\\CA=b=\sqrt{(59/12+1)^2+(16-28/13)^2}=\frac{36\sqrt{29}}{13}\\\\AB=c=\sqrt{(-1+1)^2+(16+2)^2}=18\end{gathered}

BC=a=

(59/13+1)

2

+(28/13+2)

2

=

13

90

CA=b=

(59/12+1)

2

+(16−28/13)

2

=

13

36

29

AB=c=

(−1+1)

2

+(16+2)

2

=18

s=(a+b+c)/2 = [18 + 2√29]*9/13 = 19.918....

x_in = (a * x1 + b * x2 + c * x3) / (a+b+c) = 3/2

y_in = (a * y1 + b * y2 + c * y3) / (a+b+c) = 3