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 (,)

Now we find its area, using the determinant formula:


`                         |x1 y1 1|
A = absolute value of:  |x2 y2 1|
`                         |x3 y3 1|


`                         |  1   5  1|
A = absolute value of:  | -4   0  1|
`                         | 1|

A = absolute value of  = 

TRIANGLE's AREA = 

Now we have to find the perimeter.

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

D = } =  =  =  =  = 

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

D = } =  =  =
 =  =  =  = 


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

D = } =  =  =
 =  =  =  =  = 

So the perimeter is

 +  +  =

 +  +  =

 = 

One-half the perimeter (semiperimeter) = 

Now we can use the formula:

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

RADIUS OF INSCRIBED CIRCLE OF TRIANGLE = 
                                          
RADIUS OF INSCRIBED CIRCLE OF TRIANGLE = =  = 

RADIUS OF INSCRIBED CIRCLE OF TRIANGLE =

Here's the graph:




Edwin