InterviewSolution
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InterviewSolution
| 1. |
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|` |
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Answer» 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|`. |
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