Pls help asap

i am finding answer of this question. 

mcdvoice

m  =  tan ( [ arctan (1)  + arctan (1/7) ]  /   2 )  =    .5

Corrected

Another  way

Construct  a circle at the origin with a radius of 1

Equation   is     x^2 + y^2  = 1   (1)

Slope  of line   y = x    =  1

So 

x = y

Sub this into  (1)  for y

x^2 + x^2  = 1

2x^2 = 1

x^2 = 1/2

x= 1/sqrt 2      and y =1/sqrt 2

Construct a perpendicular  line to y = x    passing through  (1/sqrt 2, 1/sqrt 2)

The equation is   

y = (-1) (x - 1/sqrt 2) + 1/sqrt2 = 

y =  -x + 2/sqrt2

y = -x + sqrt 2       (2)

Now consider the line y = (1/7)x

Sub this into (1) for y

x^2 + (1/7 x )^2   =1

x^2 + 1/49 x^2   = 1

(50 /49) x^2   =1

x^2  = 49/50

x = 7  /sqrt  (50)

y = (1/7) (7/sqrt (50)  = 1 /sqrt (50)

Now construct a perpendicular line  to y =(1/7)x    passing through  ( 7/sqrt 50 , 1/sqrt 50)

The equation  is

y = (-7)(x - 7/sqrt50) + 1/sqrt (50)

y = -7x + 50/sqrt50

y = -7x + sqrt (50)     ( 3)

Find  the  x intersection of   (2)  and (3)

-x + sqrt 2 =  -7x + sqrt 50

6x = sqrt (50)  - sqrt 2

6x  = 5sqrt (2) - sqrt 2

6x = 4sqrt 2

x = (2/3)sqrt 2

And  y =  - (2/3)sqrt 2  + sqrt 2  =   (1/3)sqrt 2

The intersection  of these two lines  is   [ (2/3)sqrt 2  ,  (1/3)sqrt 2) ] 

The slope of the line we  want (through BD)  =   [ (1/3)sqrt 2  ]  / [(2/3)sqrt 2 ]  =   1/2   =  .5 =  m

(Which we found by calculator  !!!)