Question

if G is centroid and I is the incentre of a triangle with vertices at A(-2,-1),B(1,-1),C(1,3)find disatance GI​

Answer 1

given, A (-2, -1), B (1, -1) and C(1, 3)

AB = c = √{(1 + 2)² + (-1+1)²} = 3

BC = a = √{(1 - 1)² + (3+1)²} = 4

CA = b = √{(1 + 2)² + (3 + 1)²} = 5

now formula to find incentre is ..

$\left(\frac{ax_1+bx_2+cx_3}{a+b+c}\right),\left(\frac{ay_1+by_2+cy_3}{a+b+c}\right)$

so, I = [(4 × -2 + 5 × 1 + 3 × 1)/(3 + 4 + 5), (4 × -1 + 5 × -1 + 3 × 3)/(3 + 4 + 5)]

= (0, 0)

and centroid can be found by, $\left(\frac{x_1+x_2+x_3}{3}\right),\left(\frac{y_1+y_2+y_3}{3}\right)$

= [ (-2 + 1 + 1)/3 , (-1 - 1 + 3)/3 ]

= (0, 1/3)

so, Distance GI = √{(0 - 0)² + (0 - 1/3)²} = 1/3 unit .