Question

Consider $3$ 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 $ABC$ and sum of the squares of the tangent to the interior angle can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are relatively prime numbers, then the value of $p+q$ is

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

Answer 1

The correct option is D $465$

Consider $m_1$ be the tangent of the angle between the lines
$5x-y=-4$
$3x-y=-5$
Hence
$m_1=\left|\frac{5-3}{1+15}\right|$
$m_1=\frac{1}{8}$ ...(we take the positive value of the tan of the angle)
Similarly
Consider $m_2$ be the tangent of the angle between the lines
$x+y=-8$
$3x-y=-5$
Hence
$m_2=\left|\frac{3+1}{1-3}\right|$
$m_2=2$ ...(we take the positive value of the tan of the angle)
Consider $m_3$ be the tangent of the angle between the lines
$x+y=-8$
$5x-y=-4$
Hence
$m_3=\left|\frac{5+1}{1-5}\right|$
$m_3=\frac{3}{2}$ ...(we take the positive value of the tan of the angle)
Therefore $m_3^2+m_2^2+m_1^2$
$={\left(\frac{3}{2}\right)}^2+4+\frac{1}{64}$

$=\frac{401}{64}$

Hence $p+q$
$=401+64$
$=465$