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Exercise 4.3 · Q36

Q.Find the joint equation of the pair of lines through the point (2,−1)(2, -1) and parallel to lines represented by 2x2+3xy−9y2=02x^2 + 3xy - 9y^2 = 0.

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Solving 2x2+3xy−9y2=02x^2+3xy-9y^2=0 as a quadratic in xx gives x=3y2x=\tfrac{3y}{2} or x=−3yx=-3y, i.e. slopes 23\tfrac23 and −13-\tfrac13 (lines 2x−3y=02x-3y=0, x+3y=0x+3y=0). Parallel lines through (2,−1)(2,-1): slope 23\tfrac23 gives 2x−3y−7=02x-3y-7=0; slope −13-\tfrac13 gives x+3y+1=0x+3y+1=0. Combined equation: (2x−3y−7)(x+3y+1)=0(2x-3y-7)(x+3y+1)=0, expanding to 2x2+3xy−9y2−5x−24y−7=02x^2+3xy-9y^2-5x-24y-7=0.

✓Final answer

2x2+3xy−9y2−5x−24y−7=02x^2 + 3xy - 9y^2 - 5x - 24y - 7 = 0

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