Question

If two of the lines represented by the equation $a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3=0$ are at right angles, then $a^2+d^2+ac+bd$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. $ab+bd$

Answer 1

The correct option is B $0$

${ax}^3+{bx}^{2}y+{cxy}^2+{dy}^3=0$ -------(1)
This is homogeneous equation of third degree in $x$ and $y$.
Hence represent combined equation of three straight lines passing through origin
Divide (1) by $x^3$
$\Rightarrow a+b\left(\frac{y}{x}\right)+c{\left(\frac{y}{x}\right)}^2+d{\left(\frac{y}{x}\right)}^3=0$
Substitute $\left(\frac{y}{x}\right)=m$
$\Rightarrow a+bm+{cm}^2+{dm}^3=0$
This is a cubic equation in ${}^{\prime }{m}^{\prime }$ with three roots $m_1,m_2$ and $m_3$
$\therefore m_1{m}_2{m}_3=\frac{-a}{d},m_1+m_2+m_3=-\frac{c}{d}$
and $m_1{m}_2+m_2{m}_3+m_1{m}_2=\frac{b}{d}$
For any of two lines to be perpendicular to each other i.e. $m_1{m}_2=-1$
$\therefore m_3=\frac{a}{d},m_1+m_2=-1\left(\frac{a+c}{d}\right)$
and $m_3(m_1+m_2)=\frac{b+d}{d}$
$\Rightarrow -a^2-ac=bd+d^2\Rightarrow a^2+ac+bd+d^2=0$