Show that the condition that two of the three lines represented by `ax^3+bx^2y+cxy^2+dy^3=0` may be at right angles is `a^2+ac+bd+d^2=0` .

Options

A. `a^(2)+ac+bd-d^(2)=0`
B. `a^(2)+ac-bd+d^(2)=0`
C. `a^(2)-ac+bd+d^(2)=0`
D. `a^(2)+ac+bd+d^(2)=0`

Answer

D

Explanation

The equation `ax^(3)+bx^(2)y+cxy^(2)+dy^(3)=0` is a homogeneous equation of second degree. So, it represents three straight lines passing through the origin.
Let one of the lines be `y=mx`. Then,
`ax^(2)+bx^(2)(mx)+cx(m^(2)x^(2))+d(mx)^(3)=0`
`rArr" "dm^(3)+cm^(2)+bm+a=0" ...(i)"`
This is cubic equation in m. So, it given three value of m. Let the values be `m_(1),m_(2),m_(3)`. Then,
`m_(1)+m_(2)+m_(3)=-(c)/(d)" ...(ii)"`
`m_(1)m_(2)+m_(2)m_(3)+m_(3)m_(1)=(b)/(d)" ...(iii)"`
`m_(1)m_(2)m_(3)=-(a)/(d)" ...(iv)"`
It is given that two of the three lines are at right angle.
`therefore" "m_(1)m_(2)=-1`
Putting `m_(1)m_(2)=-1` in (iv), we get `m_(3)=(a)/(d)`
Since `m_(3)` is a root of equation (i), Therefore,
`d((a)/(d))^(3)+c((a)/(d))^(2)+b((a)/(d))+a=0`
`rArr" "da^(3)+ca^(2)d+bad^(2)+ad^(3)=0 rArr a^(2)+ac+bd+d^(2)=0`