Question

Find the incentre of the triangle whose vertices are A(1, 2, 3), B(1, 5, 3) and C(5, 2, 3).

• A. (3,3,3)
  
• B. (3,2,3)
  
• C. (2, 3, 3)
• D. (3,2,2)
• E. (2,2,3)
  

Answer 1

The correct option is C (2, 3, 3)



We saw that incentre of the above triangle is given by
$l=(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c},\frac{a{z}_1+b{z}_2+c{z}_3}{a+b+c}).....\left(1\right)$
a, b and c are lengths of sides.
a is the distance between B(1, 5, 3) and C(5, 2, 3).
Using distance formula, we get a = $\sqrt{(1-5{)}^2+(5-2{)}^2+(3-3{)}^2}$
= 5 units
SImilarly we get b = 4 and c = 3 units
Substituting a = 5, b = 4 , c = 3 and the coordinates of A, B and C in (1), we get
$l=(\frac{5\left(1\right)+4\left(1\right)+3\left(5\right)}{5+4+3},\frac{5\left(2\right)+4\left(5\right)+3\left(2\right)}{5+4+3},\frac{5\left(3\right)+4\left(3\right)+3\left(3\right)}{5+4+3})$
$=(2,3,3)$