Question:
if (0,1) , (1,1) and (1,0) are middle points of the sides of a triangle then find its in centre
Answer:
Let A(x1 , y1 ), B(x1 , y1 ) and C(x1 , y1 ) be the vertices of a triangle.
x1 + x2 = 0, x2 + x3 = 2, x3 + x1 = 2
adding these, we get
2(x1 + x2 + x3 ) = 0 + 2 + 2
=> 2(x1 + x2 + x3 ) = 4
=> x1 + x2 + x3 = 2
y1 + y2 = 2, y2 + y3 = 2, y3 + y1 = 0
So, y1 + y2 + y3 = 2
Solving these equations, we get A(0, 0), B(0, 2) and C(2, 0)
Now, a = BC = 2√ 2, b = CA = 2 and c = AB = 2
So, incenter of the triangle
= {(ax1 + bx2 + cx3 )/(a + b + c), (ay1 + by2 + cy3 )/(a + b + c)}
By putting all values we get,
Incentre of the triangle ABC is (2√2, 2√2)
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