Question

The straight lines whose dc’s are given by the $al+bm+cn=0$ and $fmn+gnl+hlm=0$ are perpendicular if

• A. $\frac{f^{\prime }}{a}+\frac{g}{b}+\frac{h}{c}=0$
• B. $\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=0$
• C. $\frac{f}{a^2}+\frac{g}{b^2}+\frac{h}{c^2}=0$
• D. $\frac{a^2}{f}+\frac{b^2}{g}+\frac{c^2}{h}=0$

Answer 1

Answer: (A)

From the given equation, we have
$ag{\left(\frac{l}{m}\right)}^2+\left(af+bg-ch\right)\left(\frac{l}{m}\right)+bf=0$
or $ag{X}^2+\left(af+bg-ch\right)X+bf=0\left(X=l/m\right)$
Let two value of $l/m$ are $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$
$\therefore$ Product of roots $=\frac{l_1{l}_2}{m_1{m}_2}=\frac{bf}{ag}$
$\therefore \frac{l_1{l}_2}{\frac{f}{a}}=\frac{m_1{m}_2}{\frac{g}{b}}=\frac{n_1{n}_2}{\frac{h}{c}}$, using symmetry
Now lines to be $\perp$ if $l_1{l}_2+m_1{m}_2+n_1{n}_2=0$
$\Rightarrow k\left(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}\right)=0$
$\Rightarrow \frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$