3x² -11xy -20y² = 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: $3x^2 -11xy -20y^2 = 0$.

Solution:

Given,

Single equation of two line is $3x^2 -11xy -20y^2 = 0$

Passing point P($x_1,y_1$) = (2,1)

Finding two separate equations of lines:

$3x^2 -11xy -20y^2 = 0$

$or, 3x^2 -(15-4)xy -20y^2 = 0$

$or, 3x^2 -15xy +4xy -20y^2 = 0$

$or, 3x(x -5y) + 4y(x -5y) = 0$

$or, (x-5y)(3x +4y) = 0$

So, the separate equations of lines represented by above single equation are (x -5y = 0) and (3x +4y = 0)

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

For line 1,

Equation : $x -5y = 0$

Slope ($m_1$) = $\frac{1}{5}$

Passing point P($x_1,y_1$) = (2,1)

Equation of line parallel to it is given by,

$y - y_1 = m_1(x - x_1)$

$or, y -1 = \frac{1}{5} (x -2)$

$or, 5( y -1)= (x -2)$

$or, 5y -5 = x -2$

$or, x -5y +5-2 = 0$

$or, x -5y +3 = 0$ is the required equation.

For line 2,

Equation: $3x +4y = 0$

Slope ($m_2$) = $-\frac{3}{4}$

Passing point P($x_1,y_1$) = (2,1)

Equation of line parallel to it is given by,

$y - y_1 = m_2(x - x_1)$

$or, y -1 = - \frac{3}{4} (x -2)$

$or, 4(y -1) = -3(x -2)$

$or, 4y -4 = -3x +6$

$or, 3x + 4y -4-6 = 0$

$or, 3x +4y -10 = 0$ is the required equation.

Therefore, the required of the straight lines are (x -5y +3= 0) and (3x +4y -10 = 0).

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