Question

Formula to find the incentre of a triangle whose vertices are given

Answer 1

Formula:

Coordinates of the incenter = ( (axa + bxb+ cxc)/P , (aya + byb + cyc)/P ) Where P = (a+b+c), a,b,c = Triangle side Length

Example:

The points of a triangle are A(-3,0), B(5,0), C(-2,4). Calculate the incircle center point, area and radius.

Given,

A = (-3,0) B = (5,0) C = (-2,4)

To Find,

Incenter Area Radius

Solution:

Step 1:

First, let us calculate the sides a,b,c of the triangle. D = √(x2 - x1)2 + (y2 - y1)2
i) Calculate the side a from the given length B(5,0) C(-2,4) a = √(-2 - 5)2 + (4 - 0)2 = 8.0622
ii) Calculate the side a from the given length A(-3,0) and C(-2,4) b = √(-2 -(-3))2 + (4 - 0)2 = 4.123
iii) Calculate the side a from the given length B(5,0) and A(-3,0) c = √(-3 -5)2 + (0 - 0)2 =8
P = (a+b+c) = (8.0622 + 4.123 + 8) = 20.185

Step 2:

Substitute the a,b,c values in the coordinates formula. C(x,y) = ( (axa + bxb + cxc)/P , (aya + byb + cyc)/P ) C(x,y) = ( (8.0622(-3) + 4.123(5) + 8(-2))/20.185 , (8.0622(0) + 4.123(0) + 8(4))/20.185 ) C(x,y) = (-0.97,1.59)

Step 3:

Calculate the area of the triangle Area K = 1/4 √(P)(a-b+c)(b-c+a)(c-a+b) Area K = 1/4 √(20.185)(8.0622-4.123+8)(4.123-8+8.0622)(8-8.0622+4.123) Area K = 15.999

Step 4:

Finally, let us calculate the radius value. r = 2K/P = 2(15.999)/20.185 = 1.585

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