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Exercise 6.4 · Q11

Q.Find pp and qq, if the following equation represents a pair of perpendicular lines: 6x2+5xy−py2+7x+qy−5=06x^2 + 5xy - py^2 + 7x + qy - 5 = 0.

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Perpendicularity fixes pp via a+b=0a+b=0; then the factorisability condition abc+2fgh−af2−bg2−ch2=0abc+2fgh-af^2-bg^2-ch^2=0 becomes a quadratic in qq.

Two conditions are in play: a+b=0a+b=0 for perpendicularity, and the determinant condition for the equation to be a genuine pair of lines at all (which pins down qq).

Step 1. Read off coefficients from 6x2+5xy−py2+7x+qy−5=06x^2+5xy-py^2+7x+qy-5=0.

a=6,h=52,b=−p,g=72,f=q2,c=−5.a=6,\quad h=\tfrac52,\quad b=-p,\quad g=\tfrac72,\quad f=\tfrac{q}{2},\quad c=-5.

Step 2. Apply the perpendicularity condition a+b=0a+b=0.

6+(−p)=0 ⟹ p=6.6+(-p)=0\ \Longrightarrow\ p=6.

Step 3. Substitute p=6p=6 (so b=−6b=-6) and apply the factorisability condition. We need abc+2fgh−af2−bg2−ch2=0abc+2fgh-af^2-bg^2-ch^2=0.

abc=(6)(−6)(−5)=180.abc=(6)(-6)(-5)=180.

2fgh=2(q2)(72)(52)=35q4.2fgh=2\left(\frac{q}{2}\right)\left(\frac72\right)\left(\frac52\right)=\frac{35q}{4}.

af2=6(q2)2=3q22,bg2=−6(72)2=−1472,ch2=−5(52)2=−1254.af^2=6\left(\frac{q}{2}\right)^2=\frac{3q^2}{2},\qquad bg^2=-6\left(\frac72\right)^2=-\frac{147}{2},\qquad ch^2=-5\left(\frac52\right)^2=-\frac{125}{4}. …

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