7x² -3xy -4y² = 0 | Point (2,1) | Pair Of Straight Lines | SciPiPupil

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).

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