Question

If two of the lines \[a{{x}^{3}}+b{{x}^{2}}y+cx{{y}^{2}}+d{{y}^{3}}=0\]$\left( a\ne 0 \right)$ make complementary angles with x axis in anticlockwise then
A) a(a-c) – d(b-d) = 0
B) d(a-c) + a(b-d) = 0
C) a(a-c) + d(b-d) = 0
D) None of these

Answer 1

Hint: For solving this problem, first we assume one of the equations of line in the above given family of lines to be y = mx. Now, satisfy this equation to obtain a cubic equation in terms of slope m. By using the fact that two of the lines are complementary to each other so the product of the slope will be 1. On using the property of the product of roots in the cubic equation, we obtain the value of possible roots. On satisfying this root, we can easily show the desired expression.

Complete step-by-step answer:
Let, y = mx be any line represented in the family of lines. We are given the expression \[a{{x}^{3}}+b{{x}^{2}}y+cx{{y}^{2}}+d{{y}^{3}}=0\]
Putting the y = mx in the above expression, we get:
\[\begin{align}
  & \Rightarrow a{{x}^{3}}+b{{x}^{2}}\left( mx \right)+cx{{\left( mx \right)}^{2}}+d{{\left( mx \right)}^{3}}=0 \\
 & \Rightarrow a{{x}^{3}}+bm{{x}^{3}}+c{{m}^{2}}{{x}^{3}}+d{{m}^{3}}{{x}^{3}}=0 \\
\end{align}\]
Taking ${{x}^{3}}$ common from the equation, we get:
$\Rightarrow a+bm+c{{m}^{2}}+d{{m}^{3}}=0$, which is a cubic equation.
Let the roots of the equations be ${{m}_{1}},{{m}_{2}},{{m}_{3}}$. Now, by using the property of product of roots in general equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$, is $\dfrac{-d}{a}$.
Therefore, we get ${{m}_{1}}{{m}_{2}}{{m}_{3}}=-\dfrac{a}{d}...\left( 1 \right)$
Let one of the equations of line having slope, ${{m}_{1}}=\tan \theta $, then slope of other line, ${{m}_{2}}=\tan \left( 90-\theta \right)$ because they are complementary angles, the sum of the angles are ${{90}^{\circ }}$.
$\therefore {{m}_{2}}=\cot \theta $ because we know that $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $.
$\therefore {{m}_{1}}{{m}_{2}}=\tan \theta \cot \theta =1$
Therefore, from (1), we get ${{m}_{3}}=-\dfrac{a}{d}$
Since ${{m}_{3}}$ is the root of the above equation, we replace m as $\dfrac{-a}{d}$
$\begin{align}
  & d\left( -\dfrac{{{a}^{3}}}{{{d}^{3}}} \right)+c\left( \dfrac{{{a}^{2}}}{{{d}^{2}}} \right)+b\left( -\dfrac{a}{d} \right)+a=0 \\
 & \Rightarrow -\dfrac{{{a}^{3}}}{{{d}^{2}}}+\dfrac{c{{a}^{2}}}{{{d}^{2}}}-\dfrac{ab}{d}+a=0 \\
 & \Rightarrow \dfrac{-{{a}^{3}}+c{{a}^{2}}-abd+a{{d}^{2}}}{{{d}^{2}}}=0 \\
 & \Rightarrow -a\left( {{a}^{2}}-ac+bd-{{d}^{2}} \right)=0 \\
 & \Rightarrow a\left( a-c \right)+d\left( b-d \right)=0 \\
\end{align}$
Therefore, If two of the lines \[a{{x}^{3}}+b{{x}^{2}}y+cx{{y}^{2}}+d{{y}^{3}}=0\]$\left( a\ne 0 \right)$ make complementary angles with x axis in anticlockwise then $a\left( a-c \right)+d\left( b-d \right)=0$.
Hence, option (B) is correct.

Note: One key step in solving this problem is the assumption of the equation of a line to be y = mx. The knowledge of complementary angles is also useful. Care must be taken while handling negative signs in powers. Students should aware of property: ${{m}_{1}}{{m}_{2}}=1$