# A triangle has corners at (3 , 2 ), (6 ,7 ), and (5 ,8 ). What is the radius of the triangle's inscribed circle?

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Feb 19, 2018

The radius of the circle is 0.581

#### Explanation:

We determine the coordinates of the center of inscribed circle by investigation of the intersection of the angular bisectors.

Let,$A \equiv \left(3 , 2\right) , B \equiv \left(6 , 7\right) , \mathmr{and} C \equiv \left(5 , 8\right)$represent the vertices of the triangle ABC

Angular bisector atA:
A-=(3,2) B-=(6,7) C-=(5,8)

Slope of line AB is ${m}_{A B} = \frac{7 - 2}{6 - 3} = \frac{5}{3}$

Slope of line AB is ${m}_{A C} = \frac{8 - 2}{5 - 3} = \frac{6}{2} = 3$

Let the slope of the bisector be m

$\frac{m - \frac{5}{3}}{1 + m \times \frac{5}{3}} = \frac{3 - m}{1 + 3 \times m}$

$= \frac{3 m - 5}{3 + 5 m} = \frac{3 - m}{1 + 3 m}$

Cross multiplying

$\left(1 + 3 m\right) \left(3 m - 5\right) = \left(3 - m\right) \left(3 + 5 m\right)$

$3 m - 5 + 9 {m}^{2} - 15 m = 9 - 3 m + 15 m - 5 {m}^{2}$

$9 {m}^{2} + 5 {m}^{2} + 3 m - 15 m + 3 m - 15 m - 5 - 9 = 0$

$14 {m}^{2} - 24 m - 14 = 0$

Dividing by 14

${m}^{2} - \frac{12}{7} m - 1 = 0$

$m = 2.17 , m = - 0.46$

Equation of the bisector through A(3,2) is

$\frac{y - 2}{x - 3} = 2.17$

Simplifying

$y - 2 = 2.17 \left(x - 3\right)$

$y - 2 = 2.17 x - 6.51$

$y = 2.17 x - 4.51$

Angular bisector atB:
A-=(3,2)
B-=(6,7)
C-=(5,8)

Slope of line AB is ${m}_{A B} = \frac{7 - 2}{6 - 3} = \frac{5}{3}$

Slope of line BC is ${m}_{B C} = \frac{8 - 7}{5 - 6} = \frac{1}{-} 1 = - 1$

Let the slope of the bisector be m

$\frac{m - \frac{5}{3}}{1 + m \times \frac{5}{3}} = \frac{- 1 - m}{1 + \left(- 1\right) \times m}$

$= \frac{3 m - 5}{3 + 5 m} = \frac{- 1 - m}{1 - m}$

Cross multiplying

$\left(1 - m\right) \left(3 m - 5\right) = \left(- 1 - m\right) \left(3 + 5 m\right)$

$3 m - 5 - 3 {m}^{2} + 5 m = - 3 - 5 m - 3 m - 5 {m}^{2}$

$5 {m}^{2} - 3 {m}^{2} + 5 m + 3 m + 5 m + 3 m + 3 - 5 = 0$

$2 {m}^{2} + 16 m - 2 = 0$

Dividing by 2

${m}^{2} + 8 m - 1 = 0$

$m = 0.12 , m = - 8.12$

Equation of the bisector through B(6,7) is

$\frac{y - 7}{x - 6} = 0.12$

Simplifying

$y - 7 = 0.12 \left(x - 6\right)$

$y - 7 = 0.12 x - 0.75$

$y = 0.12 x + 6.25$

The lines $y = 2.17 x - 4.51$ and $y = 0.12 x + 6.25$ intersect at the center of the incircle

Equating rhs

$2.17 x - 4.51 = 0.12 x + 6.25$

$2.17 x - 0.12 x = 6.25 + 4.51$

$2.05 x = 10.76$

$x = \frac{10.76}{2.05}$

$x = 5.25$

$y = 2.17 x - 4.51$

$y = 2.17 \times 5.25 - 4.51$

$y = 11.39 - 4.51$

$y = 6.88$

$y = 0.12 x + 6.25$

$y = 0.12 \times 5.25 + 6.25 = 0.63 + 6.25$

$y = 6.88$

Verified

The coordinates of the centre of incircle is

$O \equiv \left(5.25 , 6.88\right)$

Slope of line AB

${m}_{A B} = \frac{5}{3}$

Point A is $A \equiv \left(3 , 2\right)$

Equation of the line AB is

$\frac{y - 2}{x - 3} = \frac{5}{3}$

$3 \left(y - 2\right) = 5 \left(x - 3\right)$

$3 y - 6 = 5 x - 15$

$5 x - 3 y + 6 - 15 = 0$

$5 x - 3 y - 9 = 0$

Center is $O \equiv \left(5.25 , 6.88\right)$

Tangent is $5 x - 3 y - 9 = 0$

The distance from O to the line AB is

given by

$| \frac{5 \times 5.25 - 3 \times 6.88 - 9}{\sqrt{{5}^{2} + {\left(- 3\right)}^{2}}} | = | \frac{- 3.39}{5.83} | = 0.581$

The radius of the circle is 0.581

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