Find the tangent of the angle between the lines whose intercepts n the axes are respectively `a ,-b` and `b ,-a`

Options

A. `(a^2-b^2)/(ab)`
B. `(b^2-a^2)/(2)`
C. `(b^2-a^2)/(2ab)`
D. None of these

Answer

C

Explanation

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)`