Find the joint equation of the pair of lines passing through the origin which are perpendicular respectively to the lines represented by 5x^{2} +2xy- 3y^{2} = 0.

#### Solution 1

Comparing the equation `5x^2 + 2xy –3y^2 = 0`, we get,

`a = 5, 2h = +2, b = –3`

Let m_{1} and m_{2} be the slopes of the lines represented by `5x^2 + 2xy –3y^2 = 0`

`m_1+m_2=-(2h)/b=-2/(-3)=2/3` ......(1)

`m_1m_2=a/b=5/(-3)`

Now required lines are perpendicular to these lines

their slopes are `-1/m_1 and -1/m_2`

Since these lines are passing through the origin, their separate equations are

`y=-1/m_1x and y=-1/m_2x`

`therefore m_1y=-x and m_2y=-x`

`x+m_1y=0 and x+m_2y=0`

their combined equation is

`(x+m_1y)(x+m_2y)=0`

`x^2+(m_1+m_2)xy+m_1m_2y^2=0`

`x^2+2/3xy+(-5)/3y^2=0`

`3x^2+2xy-5y^2=0`

#### Solution 2

Given homogeneous equation is

5x^{2} + 2xy - 3y^{2} = 0

Which is factorisable

5x^{2} + 5xy - 3xy - 3y^{2} = 0

5x(x + y) - 3y(x + y ) = 0

(x + y)(5x - 3y) = 0

∴ x + y = 0 and 5x - 3y = 0 are the two lines represented by the given equation

⇒Their slopes are -1 and `5/3`

Required two lines are respectively perpendicular to these lines.

∴ Slopes of required lines are 1 and 3/5 and the lines pass thought origin

∴ Their individual equations are

y = 1.x and `y = -3/5 x`

i.e x - y = 0 and 3x + 5y = 0

∴ Their joint equation is

(x - y) (3x + 5y) = 0

`3x^2 - 3xy + 5xy - 5y^2 = 0`

`3x^2 + 2xy - 5y^2 = 0`