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Find the Coordinates of the Centre of the Circle Inscribed in a Triangle Whose Vertices Are (−36, 7), (20, 7) and (0, −8).

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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).

Sum
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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,
\[a = BC = \sqrt{\left( 20 - 0 \right)^2 + \left( 7 + 8 \right)^2} = 25\]

\[b = AC = \sqrt{\left( 0 + 36 \right)^2 + \left( - 8 - 7 \right)^2} = 39\]
\[c = AB = \sqrt{\left( 20 + 36 \right)^2 + \left( 7 - 7 \right)^2} = 56\]

Thus, the coordinates of the in-centre of the given triangle are:
\[\left( \frac{25 \times \left( - 36 \right) + 39 \times 20 + 0}{25 + 39 + 56}, \frac{25 \times 7 + 39 \times 7 + 56\left( - 8 \right)}{25 + 39 + 56} \right)\]= \[\left( \frac{- 120}{120}, 0 \right)\]

\[\left( - 1, 0 \right)\]

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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Brief Review of Cartesian System of Rectanglar Co-ordinates
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Chapter 22: Brief review of cartesian system of rectangular co-ordinates - Exercise 22.1 [Page 13]

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R.D. Sharma Mathematics [English] Class 11
Chapter 22 Brief review of cartesian system of rectangular co-ordinates
Exercise 22.1 | Q 5 | Page 13

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