Question: Find the equation of the straight lines passing through the point (2,1) and parallel to the lines represented by: $x^2 +3xy +2y^2 = 0$.

Solution:
Given,

Single equation of two line is $x^2 +3xy +2y^2 = 0$
Passing point P($x_1,y_1$) = (2,1)

Finding two separate equations of lines:
$x^2 +3xy +2y^2 = 0$
$or, x^2 +(2+1)xy +2y^2 = 0$
$or, x^2 +2xy +xy +2y^2 = 0$
$or, x(x +2y) + y(x +2y) = 0$
$or, (x +y)(x +2y) = 0$

So, the separate equations of lines represented by above single equation are (x +y = 0) and (x +2y = 0)

[Any two straight lines, if parallel have equal slopes.]

For line 1,
Equation : $x +y = 0$
Slope ($m_1$) = $- 1$
Passing point P($x_1,y_1$) = (2,1)
Equation of line parallel to it is given by,
$y - y_1 = m_1(x - x_1)$
$or, y -1 = - 1(x -2)$
$or, y -1= -x +2$
$or, x + y -2 -1 = 0$
$or, x + y -3 = 0$ is the required equation.

For line 2,
Equation: $x +2y = 0$
Slope ($m_2$) = $-\frac{1}{2}$
Passing point P($x_1,y_1$) = (2,1)
Equation of line parallel to it is given by,
$y - y_1 = m_2(x - x_1)$
$or, y -1 = - \frac{1}{2} (x -2)$
$or, 2(y -1) = -(x -2)$
$or, 2y -2 = 2-x$
$or, x + 2y -2-2 = 0$
$or, x +2y -4 = 0$ is the required equation.

Therefore, the required of the straight lines are (x +y -3= 0) and (x +2y -4 = 0).

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