Question 86741
find the radius of the circle inscribed in the triangle bounded by the lines x-y+4=0, 7x-y-2=0 and x+y+4=0. 
actually i graph the equation and prove that x-y+4 is perpendicular to the line x+y+4. got all the points of the triangle (1,5), (-4,0) and (-1/4, -3.75), got stuck with the circle thing.. could you help me please?
<pre>
That's not quite the way to do the problem:

The formula is
                                           TRIANGLE'S AREA
RADIUS OF INSCRIBED CIRCLE OF TRIANGLE = --------------------
                                          HALF ITS PERIMETER

We first find the three corners of the triangle:

 x - y + 4 = 0
7x - y - 2 = 0

Solve that pair and get one corner point (1,5) 

 x - y + 4 = 0
 x + y + 4 = 0

Solve that pair and get the second corner point (-4,0)

7x - y - 2 = 0
 x + y + 4 = 0

Solve that pair and get the second corner point ({{{-1/4}}},{{{-15/4}}})

Now we find its area, using the determinant formula:


`                         |x<sub>1</sub> y<sub>1</sub> 1|
A = absolute value of:  {{{1/2}}}|x<sub>2</sub> y<sub>2</sub> 1|
`                         |x<sub>3</sub> y<sub>3</sub> 1|


`                         |  1   5  1|
A = absolute value of:  {{{1/2}}}| -4   0  1|
`                         |{{{-1/4}}}{{{-15/4}}} 1|

A = absolute value of {{{1/2}}}{{{(75/2)}}} = {{{75/4}}}

TRIANGLE's AREA = {{{75/4}}}

Now we have to find the perimeter.

The side of the triangle between (1,5) and (-4,0) is

D = {{{sqrt((-4-1)^2+(0-5)^2)}}}} = {{{sqrt((-5)^2 + (-5)^2)}}} = {{{sqrt(25+25)}}} = {{{sqrt(50)}}} = {{{sqrt(25*2)}}} = {{{5sqrt(2)}}}

The side of the triangle between (1,5) and (-1/4,-15/4) is

D = {{{sqrt((-1/4-1)^2+(-15/4-5)^2)}}}} = {{{sqrt((-1/4-4/4)^2 + (-15/4-20/4)^2)}}} = {{{sqrt((-5/4)^2+(-35/4)^2)}}} =
{{{sqrt(25/16+1225/16)}}} = {{{sqrt(1250/16)}}} = {{{sqrt(1250)/4}}} = {{{sqrt(625*2)/4}}} = {{{25sqrt(2)/4}}}


The side of the triangle between (-4,0) and (-1/4,-15/4) is

D = {{{sqrt((-1/4-(-4))^2+(-15/4-0)^2)}}}} = {{{sqrt((-1/4+4)^2 + (-15/4)^2)}}} = {{{sqrt((-1/4+16/4)^2+(-15/4)^2)}}} =
{{{sqrt((15/4)^2+225/16)}}} = {{{sqrt(225/16 + 225/16)}}} = {{{sqrt(450/16)}}} = {{{sqrt(450)/4}}} = {{{sqrt(225*2)/4}}} = {{{15sqrt(2)/4}}}

So the perimeter is

{{{5sqrt(2)}}} + {{{25sqrt(2)/4}}} + {{{15sqrt(2)/4}}} =

{{{20sqrt(2)/4}}} + {{{25sqrt(2)/4}}} + {{{15sqrt(2)/4}}} =

{{{60sqrt(2)/4}}} = {{{15sqrt(2)}}}

One-half the perimeter (semiperimeter) = {{{15sqrt(2)/2}}}

Now we can use the formula:

                                           TRIANGLE'S AREA
RADIUS OF INSCRIBED CIRCLE OF TRIANGLE = --------------------
                                          HALF ITS PERIMETER

RADIUS OF INSCRIBED CIRCLE OF TRIANGLE ={{{(75/4)/(15sqrt(2)/2)}}} 
                                          
RADIUS OF INSCRIBED CIRCLE OF TRIANGLE ={{{(75/4)(2/15sqrt(2))}}} = {{{5/(2sqrt(2))}}} = {{{(5sqrt(2))/(2sqrt(2)sqrt(2))}}}

RADIUS OF INSCRIBED CIRCLE OF TRIANGLE ={{{5sqrt(2)/4}}}

Here's the graph:

{{{drawing(600,600,-7,7,-7,7, graph(600,600,-7,7,-7,7,x+4,7x-2,-x-4),
   circle(-1.5,0,5sqrt(2)/4) )}}}


Edwin</pre>