Consider three lines as follows. `L_1:5x-y+4=0` `L_2:3x-y+5=0` `L_3: x+y+8=0` If these lines enclose a triangle `A B C` and the sum of the squares of the tangent to the interior angles can be expressed in the form `p/q ,` where `pa n dq` are relatively prime numbers, then the value of `p+q` is

Options

A. 500
B. 450
C. 230
D. 465

Answer

D

Explanation

Arranging the lines in desceding order of slope, we have
`m_1=5,m_2=3,and m_3=-1`
`therefore tan A =(m_1-m_2)/(1+m_1m_2)=(2)/(1+15)=(1)/(8)`
`tan B=(m_1-m_3)/(1+m_2m_3)=(3+1)/(1-3)=-2`
`tan C=(m_3-m_1)/(1+m_3m_1)=(-1-5)/(1-5)=(3)/(2)`
`Sigma tan^(2)A=(1)/(64)+4+(9)/(4)=(1+256+144)/(64)=(401)/(64)`
or `p+q=465`