find the incentre of the triangle formed by the straight line x+1=0,3x-4y=5,5x+12y=27
Answers
Answer:
Step-by-step explanation:
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
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