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

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

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Answer 1 · 2018-02-19T13:44:37

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, $A \equiv (3, 2), B \equiv (6, 7), and C \equiv (5, 8)$ $A-=(3,2), B-=(6,7), and C-=(5,8)$ represent the vertices of the triangle ABC

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

Slope of line AB is $m_{A B} = \frac{7 - 2}{6 - 3} = \frac{5}{3}$ $m_(AB)=(7-2)/(6-3)=5/3$

Slope of line AB is $m_{A C} = \frac{8 - 2}{5 - 3} = \frac{6}{2} = 3$ $m_(AC)=(8-2)/(5-3)=6/2=3$

Let the slope of the bisector be m

$\frac{m - \frac{5}{3}}{1 + m \times \frac{5}{3}} = \frac{3 - m}{1 + 3 \times m}$ $(m-5/3)/(1+mxx5/3)=(3-m)/(1+3xxm)$

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

Cross multiplying

$(1 + 3 m) (3 m - 5) = (3 - m) (3 + 5 m)$ $(1+3m)(3m-5)=(3-m)(3+5m)$

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

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

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

Dividing by 14

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

$m = 2.17, m = - 0.46$ $m=2.17, m=-0.46$

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

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

Simplifying

$y - 2 = 2.17 (x - 3)$ $y-2=2.17(x-3)$

$y - 2 = 2.17 x - 6.51$ $y-2=2.17x-6.51$

$y = 2.17 x - 4.51$ $y=2.17x-4.51$

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

Slope of line AB is $m_{A B} = \frac{7 - 2}{6 - 3} = \frac{5}{3}$ $m_(AB)=(7-2)/(6-3)=5/3$

Slope of line BC is $m_{B C} = \frac{8 - 7}{5 - 6} = \frac{1}{- 1} = - 1$ $m_(BC)=(8-7)/(5-6)=1/-1=-1$

Let the slope of the bisector be m

$\frac{m - \frac{5}{3}}{1 + m \times \frac{5}{3}} = \frac{- 1 - m}{1 + (- 1) \times m}$ $(m-5/3)/(1+mxx5/3)=(-1-m)/(1+(-1)xxm)$

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

Cross multiplying

$(1 - m) (3 m - 5) = (- 1 - m) (3 + 5 m)$ $(1-m)(3m-5)=(-1-m)(3+5m)$

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

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

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

Dividing by 2

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

$m = 0.12, m = - 8.12$ $m=0.12, m=-8.12$

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

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

Simplifying

$y - 7 = 0.12 (x - 6)$ $y-7=0.12(x-6)$

$y - 7 = 0.12 x - 0.75$ $y-7=0.12x-0.75$

$y = 0.12 x + 6.25$ $y=0.12x+6.25$

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

Equating rhs

$2.17 x - 4.51 = 0.12 x + 6.25$ $2.17x-4.51=0.12x+6.25$

$2.17 x - 0.12 x = 6.25 + 4.51$ $2.17x-0.12x=6.25+4.51$

$2.05 x = 10.76$ $2.05x=10.76$

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

$x = 5.25$ $x=5.25$

$y = 2.17 x - 4.51$ $y=2.17x-4.51$

$y = 2.17 \times 5.25 - 4.51$ $y=2.17xx5.25-4.51$

$y = 11.39 - 4.51$ $y=11.39-4.51$

$y = 6.88$ $y=6.88$

$y = 0.12 x + 6.25$ $y=0.12x+6.25$

$y = 0.12 \times 5.25 + 6.25 = 0.63 + 6.25$ $y=0.12xx5.25+6.25=0.63+6.25$

$y = 6.88$ $y=6.88$

Verified

The coordinates of the centre of incircle is

$O \equiv (5.25, 6.88)$ $O-=(5.25,6.88)$

Slope of line AB

$m_{A B} = \frac{5}{3}$ $m_(AB)=5/3$

Point A is $A \equiv (3, 2)$ $A-=(3,2)$

Equation of the line AB is

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

$3 (y - 2) = 5 (x - 3)$ $3(y-2)=5(x-3)$

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

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

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

Center is $O \equiv (5.25, 6.88)$ $O-=(5.25,6.88)$

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

The distance from O to the line AB is

given by

$∣ ∣ ∣ ∣ ∣ \frac{5 \times 5.25 - 3 \times 6.88 - 9}{\sqrt{5^{2} + {(- 3)}^{2}}} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ \frac{- 3.39}{5.83} ∣ ∣ ∣ = 0.581$ $|(5xx5.25-3xx6.88-9)/sqrt(5^2+(-3)^2)|=|(-3.39)/5.83|=0.581$

The radius of the circle is 0.581

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