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Question : 33 of 160
Marks:
+1,
-0
Solution:
We have,
AB=p+q+r AC=s+3+4 ⇒CA=−s−3−4 ‌ and ‌CB=3+−2 Area of
â–³ABC is
5√6 ⇒‌‌‌|CA×CB|=5√6 . . . (i)
Now,
CA×CB=|| =(6+4)−(2s+12)+(−s+9) =10−(2s+12)+(−s+9) ∴‌√100+(2s+12)2+(−s+9)2=5√6 ⇒√100+(4s2+144+48s)+(s2+81−18s)=10√6 ⇒‌‌325+5s2+30s=600 ⇒‌‌5s2+30s−275=0 ⇒‌‌s2+6s−55=0 ⇒‌‌s2+11s−5s−55=0 ⇒‌‌s(s+11)−5(s+11)=0 ⇒‌‌(s+11)(s−5)=0 ⇒‌‌s−5=0 or
s+11≠0 ⇒ s=5 Now,
‌‌AB=p+q+ BC=−3−+2 and
‌‌CA=−s−3−4 By applying triangle law.
AB+BC+CA=O ⇒(p+q+r)+(−3−+2)+(−5−3−4) =0 ⇒(p−8)+(q−4)+(r−2)=0 ⇒‌‌p=8,q=4 and
r=2
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