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

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).

#SciPiPupil