# Find the joint equation of the pair of lines passing through the origin which are perpendicular respectively to the lines represented by 5x^2 +2xy- 3y^2 = 0. - Mathematics and Statistics

Sum

Find the joint equation of the pair of lines passing through the origin which are perpendicular respectively to the lines represented by 5x2 +2xy- 3y2 = 0.

#### Solution 1

Comparing the equation 5x^2 + 2xy –3y^2 = 0, we get,

a = 5, 2h = +2, b = –3

Let m1 and m2 be the slopes of the lines represented by 5x^2 + 2xy –3y^2 = 0

m_1+m_2=-(2h)/b=-2/(-3)=2/3                       ......(1)

m_1m_2=a/b=5/(-3)

Now required lines are perpendicular to these lines

their slopes are -1/m_1 and -1/m_2

Since these lines are passing through the origin, their separate equations are

y=-1/m_1x and y=-1/m_2x

therefore m_1y=-x and m_2y=-x

x+m_1y=0 and x+m_2y=0

their combined equation is

(x+m_1y)(x+m_2y)=0

x^2+(m_1+m_2)xy+m_1m_2y^2=0

x^2+2/3xy+(-5)/3y^2=0

3x^2+2xy-5y^2=0

#### Solution 2

Given homogeneous equation is

5x2 + 2xy - 3y2 = 0

Which is factorisable

5x2 + 5xy - 3xy - 3y2 = 0

5x(x + y) - 3y(x + y ) = 0

(x + y)(5x - 3y) = 0

∴ x + y = 0 and 5x - 3y = 0 are the two lines represented by the given equation

⇒Their slopes are -1 and 5/3

Required two lines are respectively perpendicular to these lines.

∴ Slopes of  required lines are 1 and 3/5 and the lines pass thought origin

∴ Their individual equations are

y = 1.x and y = -3/5 x

i.e x - y = 0 and 3x + 5y = 0

∴ Their joint equation is

(x  - y) (3x + 5y) = 0

3x^2 - 3xy + 5xy - 5y^2 = 0

3x^2 + 2xy - 5y^2 = 0

Concept: Pair of Straight Lines - Pair of Lines Passing Through Origin - Homogenous Equation
Is there an error in this question or solution?

Share