Question: Find the equation of lines which pass through the point (1,2) and perpendicular to the lines represented by x²-5xy+4y²=0.

Solution:
Given,

Single equation of lines is $x^2 -5xy +4y^2 = 0$
Point P($x_1,y_1$) = (1,2)

Finding separate equations of lines:
$x^2 -5xy +4y^2 = 0$
$or, x^2 -(1+4)xy +4y^2 = 0$
$or, x^2 - xy -4xy +4y^2 = 0$
$or, x(x -y) -4y(x -y) = 0$
$or, (x -4y)(x -y) = 0$

So, the separate equations of straight lines are (x -4y = 0) and (x -y = 0).

For line 1,
Equation: (x -4y = 0)
Slope ($m_1$) = $\frac{1}{4}$
Slope of perpendicular line ($m_2$) = $-4$
Point P($x_1,y_1$) = (1,2)
Equation of perpendicular line is given by:
$y - y_1 = m_2(x -x_1)$
$or, y -2 = -4(x -1)$
$or, y -2 = -4x +4$
$or, 4x +y -2-4 = 0$
$or, 4x +y -6 = 0$ is the required equation.

For line 2,
Equation: (x -y= 0)
Slope ($m_1$) = $1$
Slope of perpendicular line ($m_2$) = $-1$
Point P($x_1,y_1$) = (1,2)
Equation of perpendicular line is given by:
$y - y_1 = m_2(x -x_1)$
$or, y -2 = -1(x -1)$
$or, y -2 = 1-x$
$or, x + y -2-1 = 0$
$or, x + y -3 = 0$ is the required equation.


Therefore, the required equations of the straight lines are (4x +y -6 = 0) and (x + y -3 = 0).

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