Find the equation of lines which pass through the point (1,0) and perpendicular to the lines represented by x²-xy-2y²=0. | SciPiPupil

Question: Find the equation of lines which pass through the point (1,0) and perpendicular to the lines represented by x²-xy-2y²=0.

Solution:

Given,

Single equation of lines is $x^2 -xy -2y^2 = 0$

Point P($x_1,y_1$) = (1,0)

Finding separate equations of lines:

$x^2 -xy -2y^2 = 0$

$or, x^2 -(2-1)xy -2y^2 = 0$

$or, x^2 - 2xy +xy -2y^2 = 0$

$or, x(x -2y) +y(x -2y) = 0$

$or, (x -2y)(x +y) = 0$

So, the separate equations of straight lines are (x -2y = 0) and (x +y = 0).

For line 1,

Equation: (x -2y = 0)

Slope ($m_1$) = $\frac{1}{2}$

Slope of perpendicular line ($m_2$) = $-2$

Point P($x_1,y_1$) = (1,0)

Equation of perpendicular line is given by:

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

$or, y -0 = -2(x -1)$

$or, y= -2x +2$

$or, 2x +y -2 = 0$ is the required equation.

For line 2,

Equation: (x +y= 0)

Slope ($m_1$) = $-1$

Slope of perpendicular line ($m_2$) = $1$

Point P($x_1,y_1$) = (1,0)

Equation of perpendicular line is given by:

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

$or, y -0 = 1(x -1)$

$or, y= x-1$

$or, x - y -1 = 0$ is the required equation.

Therefore, the required equations of the straight lines are (2x +y -2 = 0) and (x - y -1 = 0).

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