Question

The two of the straight lines represented by the equation

ax³ + bx²y + cxy² + dy³ = 0 will be at right angles, if-

• A. a² + c² = 0
• B. a² + ac + bd + d² = 0
• C. a²c² + bd + d² = 0
• D. None of these

Answer 1

Answer: (B)

a² + ac + bd + d² = 0

Answer 2

ax³ + bx²y + cxy² + dy³ = 0 … (1)

This is a homogeneous equation of third degree in x and y.

Hence represents combined equations of three straight lines passing through origin

Divide (i) by x³ Ž a + b (y/x) + c(y/x)² + d(y/x)³ = 0

Put (y/x) = m Ž a + bm + cm² + dm³ = 0.

This is a cubic equation in ‘m’ with three roots m₁, m₂, m₃ [i.e. slopes of the three lines].

$\left. \begin{array} { r } \mathrm { m } _ { 1 } \mathrm {~m} _ { 2 } \mathrm {~m} _ { 3 } = - ( \mathrm { a } / \mathrm { d } ) ; \mathrm { m } _ { 1 } + \mathrm { m } _ { 2 } + \mathrm { m } _ { 3 } = - ( \mathrm { c } / \mathrm { d } ) \\ ( * ) \\ \mathrm { m } _ { 1 } \mathrm {~m} _ { 2 } + \mathrm { m } _ { 2 } \mathrm {~m} _ { 3 } + \mathrm { m } _ { 1 } \mathrm {~m} _ { 3 } = ( \mathrm { b } / \mathrm { d } ) \end{array} \right\}$

For any of two lines to be perpendicular to each other i.e.m₁m₂ = –1.

Substituting in (*) we get

m₃ = (a/d) ; m₁ + m₂ = –1[(a + c)/d]; m₃ (m₁ + m₂)

= [(b + d)/d]

\ (a/d) [–(a + c)/d] = (b + d)/d.

Ž –a² – ac = bd + d² Ž a² + ac + bd + d² = 0.