#1**+15 **

Here's the diagram.......

The inradius will be located at (8,n) where n is the y coordinate we are looking for....let's call this "O"

The equation of the line containing segment KL is y = 0

And the equation of the line that contains segment JK is y = (15/8)x → 15x - 8y = 0

And "O" = (8,n) will be equidistant from JK and KL

Using the "formula" for the distance from a point to a line, we have

abs(15(8) - 8n) / sqrt [ 15^2 + (-8)^2 ] = abs ( n )

And we will assume that both n and 15(8) - 8n > 0....so we have

[120 - 8n] / 17 = n

[120 - 8n ] = 17n

120 = 25n

n = 120/25 = 24/5

And the distance ffrom this point to the line y = 0 is just 24/5........and this is the inradius

The equation of the incircle of triangle JKL is (x - 8)^2 + (y - 24/5)^2 = (24/5)^2

CPhill
Nov 21, 2015

#1**+15 **

Best Answer

Here's the diagram.......

The inradius will be located at (8,n) where n is the y coordinate we are looking for....let's call this "O"

The equation of the line containing segment KL is y = 0

And the equation of the line that contains segment JK is y = (15/8)x → 15x - 8y = 0

And "O" = (8,n) will be equidistant from JK and KL

Using the "formula" for the distance from a point to a line, we have

abs(15(8) - 8n) / sqrt [ 15^2 + (-8)^2 ] = abs ( n )

And we will assume that both n and 15(8) - 8n > 0....so we have

[120 - 8n] / 17 = n

[120 - 8n ] = 17n

120 = 25n

n = 120/25 = 24/5

And the distance ffrom this point to the line y = 0 is just 24/5........and this is the inradius

The equation of the incircle of triangle JKL is (x - 8)^2 + (y - 24/5)^2 = (24/5)^2

CPhill
Nov 21, 2015