The angle between the lines `xcos alpha_1+ysin alpha_1=p_1` and `xcos alpha_2+ysin alpha_2=p_2` is (A)   `|alpha_1+alpha_2|` &nbsp

Question

The angle between the lines `xcos alpha_1+ysin alpha_1=p_1` and `xcos alpha_2+ysin alpha_2=p_2` is
(A)   `|alpha_1+alpha_2|`    (B)   `|alpha_1-alpha_2|`    (C)   `|2alpha_1|`    (D)   `|2alpha_2|`   

Answer 1

Equation of line 1, `L_1 =>xcos alpha_1+ysin alpha_1=p_1`
`=>y = -x(cos alpha_1)/(sin alpha_1) +p_1/sin alpha_1`
`=> y = -cot alpha_1 x +p_1/sin alpha_1`
Comparing it with, `y = mx+c`
Slope of `L_1(m_1) = -cotalpha_1`
Equation of line 1, `L_2 =>xcos alpha_2+ysin alpha_2=p_2`
`=>y = -x(cos alpha_2)/(sin alpha_2) +p_2/sin alpha_2`
`=> y = -cot alpha_2 x +p_1/sin alpha_2`
Comparing it with, `y = mx+c`
Slope of `L_2(m_2) = -cotalpha_2`
So, `tan alpha = |(m_1-m_2)/(1+m_1m_2)|`
`tan alpha = |(cot alpha_2 - cot alpha_1)/(1+cot alpha_1cot alpha_2)|`
`=>tan alpha = |(sin alpha_1cosalpha _2 - sin alpha_2cosalpha_1)/(cos alpha_1cosalpha _2 - sin alpha_2sinalpha_1)|`
`=>tan alpha = |(sin(alpha_1-alpha_2))/(cos(alpha_1-alpha_2))|`
`=>tan alpha = |tan(alpha_1 - alpha_2)|`
`=> alpha = |alpha_1 - alpha_2|`
So, angle between these two lines will be `|alpha_1 - alpha_2|`.