InterviewSolution
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InterviewSolution
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Find the tangent of the angle between the lines whose intercepts n theaxes are respectively `a ,-b` and `b ,-a`A. `(a^2-b^2)/(ab)`B. `(b^2-a^2)/(2)`C. `(b^2-a^2)/(2ab)`D. None of these |
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Answer» Correct Answer - C Sinc, intercepts on the axes are a,-b then equation of the line is `(x)/(a)-(y)/(b)=1` `rArr(y)/(b)-(x)/(a)=1` `rArr y=(bx)/(a)-b` So, the lope of this line i.e., `m_1=(b)/(a)` Also, for intercepts on the axes as b and -a, then equation of the line is `(x)/(b)-(y)/(a)=1` `rArr(y)/(a)-(x)/(b)-1rArry=(a)/(b)x-a` and slope of this line i.e,`m_2=(a)/(b)` `tan theta =(b/a-a/b)/(1+a/b.b/a)=((b^2-a^2)/(ab))/(2)=(b^2-a^2)/(2ab)` |
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