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Question 122 of 139

Q.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=05x^2 + 2xy - 3y^2 = 0.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2017Subjective· 3mImportance★★★★★
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Factor the given pair, find perpendicular slopes, then form the new joint equation.

5x2+2xy−3y2=5x2+5xy−3xy−3y2=5x(x+y)−3y(x+y)=(5x−3y)(x+y)5x^2+2xy-3y^2 = 5x^2+5xy-3xy-3y^2 = 5x(x+y)-3y(x+y) = (5x-3y)(x+y)

So the given lines are 5x−3y=05x-3y=0 (slope 53\tfrac53) and x+y=0x+y=0 (slope −1-1).

Lines through the origin perpendicular to these have slopes −35-\tfrac35 and 11 respectively:

  • Perpendicular to 5x−3y=05x-3y=0: y=−35x⇒3x+5y=0y = -\dfrac35 x \Rightarrow 3x+5y=0
  • Perpendicular to x+y=0x+y=0: y=x⇒x−y=0y = x \Rightarrow x-y=0

Joint equation: (3x+5y)(x−y)=0(3x+5y)(x-y) = 0

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