A triangle has corners at #(4 , 5 )#, #(8 ,2 )#, and #(4 ,7 )#. What is the radius of the triangle's inscribed circle?

1 Answer
Bio
Feb 6, 2016

#r = frac{5sqrt41}{8}#

Explanation:

The first step is to find the location of the center of the circle.

It is where the perpendicular bisectors of the edges of the triangle meet. So it is the intersection of 3 lines that we are finding. But actually 2 lines is enough to find the intersection; the third line will necessarily intersect at the same point.

The perpendicular bisector is a locus of points that are equidistant from the corners of the triangle. From the Pythagoras Theorem, an expression for distance from a point with coordinate #(p,q)# looks something like this.

#sqrt{(x-p)^2+(y-q)^2}#

So to find the locus of points that are equidistant from #(4,5)# and #(4,7)#, we need to solve the following equation.

#sqrt{(x-4)^2+(y-5)^2} = sqrt{(x-4)^2+(y-7)^2}#

#(x-4)^2+(y-5)^2 = (x-4)^2+(y-7)^2#

#(y-5)^2 = (y-7)^2#

#y^2-10y+25 = y^2-14y+49#

#4y = 24#

#y = 6#

To find the locus of points that are equidistant from #(4,5)# and #(8,2)#, we need to solve the following equation.

#sqrt{(x-4)^2+(y-5)^2} = sqrt{(x-8)^2+(y-2)^2}#

#(x-4)^2+(y-5)^2 = (x-8)^2+(y-2)^2#

#(x^2 - 8x + 16) + (y^2 - 10y + 25) = (x^2 - 16x + 64) + (y^2 - 4y + 4)#

# - 8x + 16 - 10y + 25 = - 16x + 64 - 4y + 4#

# 8x = 27 + 6y#

Since we know from the first equation #y = 6#, we can substitute it directly into the second equation. So,

#x = frac{27 + 6(6)}{8} = 63/8#

The coordinates of the circumcenter is #(63/8,6)#.

The radius can be found by calculating the distance from the center of the circle to any of the corners.

#sqrt{(63/8-8)^2 + (6-2)^2} = frac{5sqrt41}{8}#

#sqrt{(63/8-4)^2 + (6-5)^2} = frac{5sqrt41}{8}#

#sqrt{(63/8-4)^2 + (6-7)^2} = frac{5sqrt41}{8}#

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