Given pair of perpendicular lines are given as, ax2+6xy+by2−10x+10y−6=0 ...(i) General equation is given as, ax2+2hxy+by2+2gx+2fy+c=0...(ii) Compare Eqs. (i) and (ii), we obtain a=a,2h=6⇒h=3, b=b, 2g=−10⇒g=−5 2f=10⇒f=5,c=−6 |
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|=0 ⇒a(−6b−25)−3(−18+25−5(15−5b)=0 ⇒25a+25b+6ab+96=0 ...(iii) Since, lines are perpendicular ⇒a+b=0 or a=−b ...(iv) Use Eq. (iv) in Eq. (iii), 25(a−a)+6a(−a)+96=0 ⇒6a2=96⇒a2=16 ⇒a=±4 ⇒|a|=4