7. The lines 36x2 – 33xy – 20y2 = 0 are equally inclined to the line
1) 3x + 11xy – 5 = 0
2) 3x – 11xy + 5 = 0
3) 3x + 4y + 5 = 0
4) 3x + 7y + 5 = 0
Answers
Question :- The lines 36x² – 33xy – 20y² = 0 are equally inclined to the line
1) 3x + 11xy – 5 = 0
2) 3x – 11xy + 5 = 0
3) 3x + 4y + 5 = 0
4) 3x + 7y + 5 = 0
Solution :-
we know that, the equation of bisector of angles between pair of lines ax² + 2hxy + by² = 0 is :-
- (x² - y²)/(a - b) = xy/h .
Comparing the given lines 36x² – 33xy – 20y² = 0 with ax² - 2hxy + by² = 0 we get :-
- a = 36
- 2h = (-33) => h = (-33/2)
- b = (-20)
Putting values we get :-
→ (x² - y²)/(a - b) = xy/h
→ (x² - y²)/(36 - (-20)) = xy/(-33/2)
→ (x² - y²)/(36 + 20) = (-2xy) / 33
→ (x² - y²)/56 = (-2xy)/33
→ 33(x² - y²) = (-112xy)
→ 33x² - 33y² = (-112xy)
→ 33x² + 112xy - 33y² = 0
→ 33x² + 121xy - 9xy - 33y² = 0
→ 11x(3x + 11y) - 3y(3x + 11y) = 0
→ (11x - 3y)(3x + 11y) = 0
Therefore, one of the bisectors is 3x + 11y = 0 which is parallel to the line 3x + 11y = 5 .
Hence, the line 3x + 11y - 5 = 0 is equally inclined to the given lines.
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and is 3x+11y-5 hop it helps
