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Question
Find the joint equation of pair of lines through the origin which is perpendicular to the lines represented by 5x2 + 2xy - 3y2 = 0
Solution 1
Comparing the equation `5x^2 + 2xy –3y^2 = 0`, we get,
`a = 5, 2h = +2, b = –3`
Let m1 and m2 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
5x2 + 2xy - 3y2 = 0
Which is factorisable
5x2 + 5xy - 3xy - 3y2 = 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`
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