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Exercise 6.4 · Q4

Q.Show that the equation 2x2−xy−3y2−6x+19y−20=02x^2 - xy - 3y^2 - 6x + 19y - 20 = 0 represents a pair of intersecting lines. Show further that the angle between them is tan⁡−1(5)\tan^{-1}(5).

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Confirm the factorisability condition abc+2fgh−af2−bg2−ch2=0abc+2fgh-af^2-bg^2-ch^2=0 so the equation is a genuine pair; since h2−ab>0h^2-ab>0 the lines are real, distinct (intersecting); the angle formula then gives tan⁡θ=5\tan\theta=5.

Step 1. Read off the coefficients. Comparing with ax2+2hxy+by2+2gx+2fy+c=0ax^2+2hxy+by^2+2gx+2fy+c=0:

a=2, h=−12, b=−3, g=−3, f=192, c=−20.a=2,\ h=-\tfrac12,\ b=-3,\ g=-3,\ f=\tfrac{19}{2},\ c=-20.

Step 2. Check the equation is a genuine pair of lines. The condition is abc+2fgh−af2−bg2−ch2=0abc+2fgh-af^2-bg^2-ch^2=0.

abc=2(−3)(−20)=120,abc=2(-3)(-20)=120,

2fgh=2(192)(−3)(−12)=572,2fgh=2\left(\tfrac{19}{2}\right)(-3)\left(-\tfrac12\right)=\tfrac{57}{2},

af2=2(192)2=3612,bg2=−3(−3)2=−27,ch2=−20(14)=−5.af^2=2\left(\tfrac{19}{2}\right)^2=\tfrac{361}{2},\qquad bg^2=-3(-3)^2=-27,\qquad ch^2=-20\left(\tfrac14\right)=-5.

Sum: 120+572−3612−(−27)−(−5)=120+28.5−180.5+27+5=0.120+\tfrac{57}{2}-\tfrac{361}{2}-(-27)-(-5)=120+28.5-180.5+27+5=0. ✓ So the equation genuinely represents a pair of straight lines.

Step 3. Check that the lines are distinct and real (hence intersecting, not parallel).

h2−ab=(−12)2−(2)(−3)=14+6=254>0.h^2-ab=\left(-\tfrac12\right)^2-(2)(-3)=\tfrac14+6=\tfrac{25}{4}>0. …

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