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Exercise 6.4 · Q13

Q.For what value of kk does the equation 12x2+2kxy+2y2+11x−5y+2=012x^2 + 2kxy + 2y^2 + 11x - 5y + 2 = 0 represent two straight lines?

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Apply the factorisability condition abc+2fgh−af2−bg2−ch2=0abc+2fgh-af^2-bg^2-ch^2=0 to get a quadratic in kk.

Step 1. Read off coefficients from 12x2+2kxy+2y2+11x−5y+2=012x^2+2kxy+2y^2+11x-5y+2=0.

a=12,h=k,b=2,g=112,f=−52,c=2.a=12,\quad h=k,\quad b=2,\quad g=\tfrac{11}{2},\quad f=-\tfrac52,\quad c=2.

Step 2. Apply the factorisability condition.

abc=12(2)(2)=48.abc=12(2)(2)=48.

2fgh=2(−52)(112)k=2(−554)k=−55k2.2fgh=2\left(-\frac52\right)\left(\frac{11}{2}\right)k=2\left(-\frac{55}{4}\right)k=-\frac{55k}{2}.

af2=12(52)2=75,bg2=2(112)2=1212,ch2=2k2.af^2=12\left(\frac52\right)^2=75,\qquad bg^2=2\left(\frac{11}{2}\right)^2=\frac{121}{2},\qquad ch^2=2k^2.

Step 3. Combine.

48−55k2−75−1212−2k2=0.48-\frac{55k}{2}-75-\frac{121}{2}-2k^2=0.

Constants: 48−75=−2748-75=-27, and −27−1212=−1752-27-\tfrac{121}{2}=-\tfrac{175}{2}. So …

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