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Question : 149 of 150
Marks:
+1,
-0
Solution:
Given equation is
x2−7xy+12y2=0On comparing with
ax2+2hxy+by2=0∴‌a=1,h=−‌,b=12∴‌m1+m2=−‌‌ and ‌‌m1m2=‌⇒m1+m2=‌ . . . (i)
Now,
‌‌(m1−m2)=√(m1+m2)2−4m1m2‌=√(‌)2−4×‌‌=√‌−‌‌=√‌ =‌√‌=‌⇒‌‌m1−m2=‌ . . . (ii)
On solving Eqs. (i) and (ii), we get
‌m1=‌‌m2=‌ ‌ Now, ‌tan‌θ‌=‌=‌tan‌θ‌=‌=‌≠0 Hence, given equation is a non-perpendicular straight lines.
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