InterviewSolution
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InterviewSolution
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What is the acute angle between the lines `Ax+By=A+B` and `A( x-y)+B(x+y)=2B` ?A. `45^(@)`B. `tan^(-1) (A/sqrt(A^(2)+B^(2)))`C.D. `60^(@)` |
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Answer» Correct Answer - A Lines are `L_(1) equiv Ax+By=A+B` and `L_(2) equiv A(x-y)+B(x+y)=2B` Slope of `L^(1)` is `- A/B` `m_(1)=-A/B" "[m_(1)" is the side of line "L_(1)]` For line `L_(2)` : `Ax-Ay+Bx+By=2B` `(A+B)x-(A-B)y=2B`. Slope of line `L_(2)` in `((A+B))/(A-B)` `m_(2)=((A+B))/((A-B))" "[m_(2)" is the slope of line "L_(2)]` If angle between line `L_(1)` and `L_(2)` is `theta` then `tan theta=(m_(1)-m_(2))/(1+m_(1)m_(2))` `=(-A/B-(A+B)/(A-B))/(1+(-A/B)xx((A+B)/(A-B)))=((-A(A-B)-B(A+B))/(B(A-B)))/((B(A-B)-A(A+B))/(B(A-B)))` `=(-A^(2)+AB-AB-B^(2))/(AB-B^(2)-A^(2)-AB)=(-B^(2)-A^(2))/(-B^(2)-A^(2))=1` so, `theta=pi/4` |
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