Question

The three sides of triangle are given by $L_r\equiv xcos{\theta }_r+ysin{\theta }_r=0(r=1,2,3)$, then the orthocenter of the triangle is given by

• A. $L_{3}sin\left({\theta }_2{\theta }_3\right)=L_{2}sin({\theta }_3-{\theta }_1)=L_{1}sin({\theta }_1-{\theta }_2)$
  
• B. $L_{1}cos({\theta }_2+{\theta }_3)=L_{2}cos({\theta }_1+{\theta }_3)=L_{3}cos({\theta }_3+{\theta }_2)$
  
• C. $L_2+\lambda L_3=0,L_3+\mu L_1=0,(\lambda ,\mu$ are arbitrary constant)
  
• D. $L_{1}cos({\theta }_2-{\theta }_3)=L_{2}cos({\theta }_2-{\theta }_1)=L_{3}cos({\theta }_1-{\theta }_2)$

Answer 1

Answer: (C), (D)

The correct options are
C $L_2+\lambda L_3=0,L_3+\mu L_1=0,(\lambda ,\mu$ are arbitrary constant)

D $L_{1}cos({\theta }_2-{\theta }_3)=L_{2}cos({\theta }_2-{\theta }_1)=L_{3}cos({\theta }_1-{\theta }_2)$

Equation of AD must be $L_2+\lambda L_3=0,\lambda$ can be determined since $AD\perp BCand\lambda =\frac{-cos({\theta }_1-{\theta }_2)}{cos({\theta }_3-{\theta }_1)}$

$\therefore$ Equation of AD is
$L_{2}cos({\theta }_3-{\theta }_1)=L_{3}cos({\theta }_1-{\theta }_2)$
We can similarly get the equation of altitude BE.