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What is pahrmacy..%owkwhwgevev |
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Answer» We have, x=py+q⇒y= p x−q ⋯(i) And z=ry+s⇒y= r z−s …(ii) ⇒ p x−q = 1 y = r z−s [ using Eqs. ( i ) and (ii)]⋯(III) Similarly, p ′
x−q ′
= 1 y = r ′
z−s ′
…(iv) From Eqs. (iii) and (iv),a 1 =p,B 1 =1,c 1 =r and a 2 =p ′ ;b 2 =1,c 2 =r ′
if these given lines are perpendicular to each other, then a 1 a 2 +b 1 b 2 +c 1 c 2 =0 ⇒pp ′ +1+rr ′ =0 Which is the required CONDITION. |
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