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The joint equation of pair of lines whic...

The joint equation of pair of lines which passes through origin and are perpendicular to the lines represented the equation `y^(2) +3xy -6x +5y - 14 = 0` will be

A

`y^(2) - 3xy = 0`

B

`3y^(2) - xy = 0`

C

`x^(2) - 3xy = 0`

D

`3x^(2) - xy = 0`

Text Solution

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To find the joint equation of the pair of lines that pass through the origin and are perpendicular to the lines represented by the equation \( y^2 + 3xy - 6x + 5y - 14 = 0 \), we can follow these steps: ### Step 1: Identify the Homogeneous Part of the Given Equation The given equation is: \[ y^2 + 3xy - 6x + 5y - 14 = 0 \] To find the pair of lines, we first consider the homogeneous part of this equation, which is obtained by ignoring the constant term (-14) and the non-homogeneous terms (-6x and +5y): \[ y^2 + 3xy = 0 \]
To find the joint equation of the pair of lines that pass through the origin and are perpendicular to the lines represented by the equation \( y^2 + 3xy - 6x + 5y - 14 = 0 \), we can follow these steps: ### Step 1: Identify the Homogeneous Part of the Given Equation The given equation is: \[ y^2 + 3xy - 6x + 5y - 14 = 0 \] To find the pair of lines, we first consider the homogeneous part of this equation, which is obtained by ignoring the constant term (-14) and the non-homogeneous terms (-6x and +5y): ...
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