Question

The equation of the lines passing through the origin and parallel to the lines represented by the equation \[2{{x}^{2}}-xy-6{{y}^{2}}+7x+21y-15=0\] , is
A. \[2{{x}^{2}}-xy-6{{y}^{2}}=0\]
B. \[6{{x}^{2}}-xy+2{{y}^{2}}=0\]
C. \[6{{x}^{2}}-xy-2{{y}^{2}}=0\]
D. \[2{{x}^{2}}+xy-6{{y}^{2}}=0\]

Answer 1

Hint: The condition for the equation of the line passing through the origin and parallel to a given line is that the value of one degree variable is zero. Thus here we have to put this condition.

Formula Used: $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$

Complete step by step solution: The given equation of the straight line is \[2{{x}^{2}}-xy-6{{y}^{2}}+7x+21y-15=0\]
The general equation of straight line is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$.
The condition for the equation of the line passing through the origin and parallel to a given line is that the value of one degree variable is zero. If general equation the one degree variable are $2gx,2fy,c$ as they contain only one coordinate or no coordinate.
Thus in the given equation \[2{{x}^{2}}-xy-6{{y}^{2}}+7x+21y-15=0\]the one degree variables are $7x,21y,-15$ . The value of this variables is zero. Thus the equation of the lie parallel and passing through the origin can be given as
\[2{{x}^{2}}-xy-6{{y}^{2}}=0\]
Thus we can write that the equation of the lines passing through the origin and parallel to the lines represented by the equation \[2{{x}^{2}}-xy-6{{y}^{2}}+7x+21y-15=0\] , is \[2{{x}^{2}}-xy-6{{y}^{2}}=0\].

Option ‘A’ is correct

Note: The condition for parallel line is that the slopes of the lines must be equal. The another condition is that the straight lines must be cut by a transversal.