Question: Find the equation of the straight lines passing through the point (2,1) and parallel to the lines represented by: $7x^2 -3xy -4y^2 = 0$.

Solution:
Given,

Single equation of two line is $7x^2 -3xy -4y^2 = 0$
Passing point P($x_1,y_1$) = (2,1)

Finding two separate equations of lines:
$7x^2 -3xy -4y^2 = 0$
$or, 7x^2 -(7-4)xy -4y^2 = 0$
$or, 7x^2 -7xy +4xy -4y^2 = 0$
$or, 7x(x -y) +4y(x -y) = 0$
$or, (7x +4y)(x -y) = 0$

So, the separate equations of lines represented by above single equation are (7x +4y = 0) and (x - y = 0).

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

For line 1,
Equation : $7x +4y = 0$
Slope ($m_1$) = $- \frac{7}{4}$
Passing point P($x_1,y_1$) = (2,1)
Equation of line parallel to it is given by,
$y - y_1 = m_1(x - x_1)$
$or, y -1 = - \frac{7}{4}(x -2)$
$or, 4(y -1) = -7(x -2)$
$or, 4y -4 = -7x +14$
$or, 7x +4y -4-14=0$
$or, 7x +4y -18 = 0$ is the required equation.

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

Therefore, the required of the straight lines are (x - y -1 = 0) and (7x +4y -18 = 0).

#SciPiPupil