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Question
Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8).
Solution
The coordinates of the in-centre of a triangle whose vertices are \[A\left( x_1 , y_1 \right), B\left( x_2 , y_2 \right)\text{ and }C\left( x_3 , y_3 \right)\] are \[\left( \frac{a x_1 + b x_2 + c x_3}{a + b + c}, \frac{a y_1 + b y_2 + c y_3}{a + b + c} \right)\], where a = BC, b = AC and c = AB.
Let A(−36, 7), B(20, 7) and C(0, −8) be the coordinates of the vertices of the given triangle.
Now,
\[a = BC = \sqrt{\left( 20 - 0 \right)^2 + \left( 7 + 8 \right)^2} = 25\]
Thus, the coordinates of the in-centre of the given triangle are:
\[\left( \frac{25 \times \left( - 36 \right) + 39 \times 20 + 0}{25 + 39 + 56}, \frac{25 \times 7 + 39 \times 7 + 56\left( - 8 \right)}{25 + 39 + 56} \right)\]= \[\left( \frac{- 120}{120}, 0 \right)\]
Hence, the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8) is \[\left( - 1, 0 \right)\]
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