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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