Question

If $\theta$ is angle between two non parallel lines then $\sin \theta$ is

• A. $\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$
• B. $\frac{2\sqrt{h^2-ab}}{\sqrt{(a-b{)}^2+4h^2}}$
• C. $\frac{2\sqrt{h^2-ab}}{\sqrt{g^2-ac}}$
• D. $\frac{2\sqrt{h^2-ab}}{f^2-bc}$

Answer 1

Answer: (B)

$\frac{2\sqrt{h^2-ab}}{\sqrt{(a-b{)}^2+4h^2}}$

If the lines

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

$a{x}^2+2hxy+b{y}^2-2gx-2fy+c=0$

represents the straight lines

The equations can be written as product of $2$ linear factors,

$(lx+my)(px+qy)$

where, $lp=a,mq=b$ & $lq+mp=2b$,

So the angle between lines is given by

$\sin \theta =\frac{\sqrt{(l^2+m^2)(p^2+q^2)-(lp-mq{)}^2}}{\sqrt{(l^2+m^2)(p^2+q^2)}}$

$=\frac{\sqrt{(l^2{p}^2+l^2{q}^2+m^2{p}^2+m^2{q}^2)-(l^2{p}^2+m^2{q}^2+2lpmq)}}{\sqrt{(lp-mq{)}^2+(lq+mp{)}^2}}$

$=\frac{\sqrt{l^2{q}^2+m_1^{2}p-2lpmq}}{\sqrt{(a-b{)}^2+4h^2}}$

$=\frac{2\sqrt{h^2-ab}}{\sqrt{(a-b{)}^2+4h^2}}$.