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Exercise 4.3 · Q39

Q.Show that equation 2x2−xy−3y2−6x+19y−20=02x^2 - xy - 3y^2 - 6x + 19y - 20 = 0 represents a pair of lines.

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Here a=2,h=−12,b=−3,g=−3,f=192,c=−20a=2,h=-\tfrac12,b=-3,g=-3,f=\tfrac{19}2,c=-20. h2−ab=14+6=254>0h^2-ab=\tfrac14+6=\tfrac{25}4>0. Determinant: abc=120abc=120, 2fgh=2(192)(−3)(−12)=5722fgh=2\left(\tfrac{19}2\right)(-3)\left(-\tfrac12\right)=\tfrac{57}2, af2=3612af^2=\tfrac{361}2, bg2=−27bg^2=-27, ch2=−5ch^2=-5. Sum: $120+\tfrac{57}2-\tfrac{361}2-(-27)-(-5)=120+ …

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