First slide

Introduction to 3-D Geometry

Question

The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 4, 0)$ is

Very Easy

A

$(1, 1, 0)$
B

$(2, 2, 0)$
C

$(3, 3, 0)$
D

$(\frac{1}{2}, \frac{1}{2}, 0)$

Solution

The incentre of the triangle formed by the points $(x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}), (x_{3}, y_{3}, z_{3})$ is $(\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}, \frac{a z_{1} + b z_{2} + c z_{3}}{a + b + c})$ where $a, b, c$ are the sides of the triangle.
The vertices are $O (0, 0, 0), A (3, 0, 0), B (0, 4, 0)$
Here the sides are $O A = 3, O B = 4, A B = 5$
Therefore, the incentre of the triangle is $(1, 1, 0)$

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