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Locus of point of intersection of the perpendicular lines one belonging to (x + y – 2) + λ(2x + 3y – 5) = 0 and other to (2x + y – 11) + λ(x + 2y – 13) = 0 is a
  • a)
    circle
  • b)
    straight line
  • c)
    pair of lines
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?
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Understanding the Problem
To find the locus of the intersection of two perpendicular lines, we start with the given equations:
1. (x + y – 2) + λ(2x + 3y – 5) = 0
2. (2x + y – 11) + μ(x + 2y – 13) = 0
Here, λ and μ are parameters that allow us to express families of lines.
Finding the Direction Ratios
- For the first line, the direction ratios can be derived from the coefficients of x and y:
- Direction ratios: (1 + 2λ, 1 + 3λ)
- For the second line:
- Direction ratios: (2 + μ, 1 + 2μ)
Condition for Perpendicularity
For the lines to be perpendicular, the dot product of their direction ratios must equal zero:
(1 + 2λ)(2 + μ) + (1 + 3λ)(1 + 2μ) = 0
This provides a relationship between λ and μ.
Finding the Intersection Point
By solving the equations for the intersection point (x, y) based on the parameters λ and μ, we can express x and y in terms of these parameters.
Locus of the Intersection Points
The locus of the intersection points, as λ and μ vary, will be a set of points that satisfy the perpendicularity condition. After simplifying the relationship derived from the above conditions, it turns out that this locus represents a conic section.
Conclusion
Through analysis and simplification, it can be shown that this locus is indeed a circle, as the derived equation corresponds to the general form of a circle in Cartesian coordinates.
Thus, the correct answer is option 'A': Circle.
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Locus of point of intersection of the perpendicular lines one belonging to (x + y – 2) + λ(2x + 3y – 5) = 0 and other to (2x + y – 11) + λ(x + 2y – 13) = 0 is aa)circleb)straight linec)pair of linesd)None of theseCorrect answer is option 'A'. Can you explain this answer?
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