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  • If two of the three lines given by  <img alt="\mathrm{a x^3+b x^2 y+c x y^2+d y^3=0(a \neq 0)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba%20x%5E3&plus;b%20x%5

If two of the three lines given by  \mathrm{a x^3+b x^2 y+c x y^2+d y^3=0(a \neq 0)}  make complementary angles with \mathrm{\mathrm{x}-}axis, then \mathrm{\mathrm{a}(\mathrm{a}-\mathrm{c})=}

Option: 1

\mathrm{d(b-d)}


Option: 2

\mathrm{d(d-b)}


Option: 3

\mathrm{b(b-d)}


Option: 4

\mathrm{b(d-b)}


Answers (1)

best_answer

Let  \mathrm{y=m_1 x, y=m_2 x}  and \mathrm{ y=m_3 x} be the lines represented by the given equation. Proceeding as in Example 8, we get \mathrm{m}_1+\mathrm{m}_2+\mathrm{m}_3=-\mathrm{c} / \mathrm{d}........(i)

\mathrm{m}_1 \mathrm{~m}_2+\mathrm{m}_2 \mathrm{~m}_3+\mathrm{m}_3 \mathrm{~m}_1=\mathrm{b} / \mathrm{d}............(ii)

and  \mathrm{m}_1 \mathrm{~m}_2 \mathrm{~m}_3=-\mathrm{a} / \mathrm{d}.................................(iii)

It is given that two of the three lines make complementary angles with \mathrm{x} -axis. So, if one line makes an angle \mathrm{\theta}  with \mathrm{x}-axis, then other makes \mathrm{ 90^{\circ}-\theta} with \mathrm{x} -axis. Let these two lines be  \mathrm{\mathrm{y}=\mathrm{m}_1 \mathrm{x}} and \mathrm{y}=\mathrm{m}_2 \mathrm{x}.

Then, \mathrm{m}_1=\tan \theta \text { and } \mathrm{m}_2=\tan \left(90^{\circ}-\theta\right) \quad \Rightarrow \mathrm{m}_1 \mathrm{~m}_2=1

putting \mathrm{m}_1 \mathrm{~m}_2=1 in (iii), we get \mathrm{m}_3=-\mathrm{a} / \mathrm{d}
since  \mathrm{y=m_3 x}  is a line represented by  \mathrm{a x^3+b x^2 y+c x y^2+d y^3=0}

\therefore \mathrm{ax}^3+\mathrm{bm}_3 \mathrm{x}^3+\mathrm{cm}_3{ }^2 \mathrm{x}^3+\mathrm{dm}_3{ }^3 \mathrm{x}^3=0 \quad \Rightarrow \mathrm{dm}_3{ }^3+\mathrm{cm}_3{ }^2+\mathrm{bm}_3+\mathrm{a}=0

\mathrm{\Rightarrow-d\left(\frac{a^3}{d^3}\right)+c\left(\frac{a^2}{d^2}\right)-\frac{a b}{d}+a=0}

\mathrm{\text { [Putting } \left.m_2=-a / d\right] \Rightarrow-a^2+a c-b d+d^2=0 \Rightarrow a(a-c)+d(b-d)=0}

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Rishabh

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