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JEE Main 2024
Straight Lines
Straight Lines and Pair of Straight Lines
Hard

Question

If x2y2+2hxy+2gx+2fy+c=0x^2-y^2+2 h x y+2 g x+2 f y+c=0 is the locus of a point, which moves such that it is always equidistant from the lines x+2y+7=0x+2 y+7=0 and 2xy+8=02 x-y+8=0, then the value of g+c+hfg+c+h-f equals

Options

Solution

Key Concepts and Formulas

  • Equation of Angle Bisectors: The equations of the angle bisectors between two lines A1x+B1y+C1=0A_1x + B_1y + C_1 = 0 and A2x+B2y+C2=0A_2x + B_2y + C_2 = 0 are given by: A1x+B1y+C1A12+B12=±A2x+B2y+C2A22+B22\frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}}
  • General Equation of a Pair of Straight Lines: The general equation of a pair of straight lines is given by: ax2+2hxy+by2+2gx+2fy+c=0ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0
  • Distance Formula: The distance of a point (x,y)(x, y) from a line Ax+By+C=0Ax + By + C = 0 is given by: d=Ax+By+CA2+B2d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}

Step-by-Step Solution

Step 1: Find the equations of the angle bisectors.

We are given the two lines: L1:x+2y+7=0L_1: x + 2y + 7 = 0 L2:2xy+8=0L_2: 2x - y + 8 = 0

Using the formula for the angle bisectors, we have: x+2y+712+22=±2xy+822+(1)2\frac{x + 2y + 7}{\sqrt{1^2 + 2^2}} = \pm \frac{2x - y + 8}{\sqrt{2^2 + (-1)^2}} x+2y+75=±2xy+85\frac{x + 2y + 7}{\sqrt{5}} = \pm \frac{2x - y + 8}{\sqrt{5}} x+2y+7=±(2xy+8)x + 2y + 7 = \pm (2x - y + 8)

Step 2: Find the two angle bisector equations.

Case 1: x+2y+7=2xy+8x + 2y + 7 = 2x - y + 8 x3y+1=0x - 3y + 1 = 0

Case 2: x+2y+7=(2xy+8)x + 2y + 7 = -(2x - y + 8) x+2y+7=2x+y8x + 2y + 7 = -2x + y - 8 3x+y+15=03x + y + 15 = 0

Step 3: Combine the angle bisector equations to form a single equation.

The combined equation of the pair of straight lines is: (x3y+1)(3x+y+15)=0(x - 3y + 1)(3x + y + 15) = 0 3x2+xy+15x9xy3y245y+3x+y+15=03x^2 + xy + 15x - 9xy - 3y^2 - 45y + 3x + y + 15 = 0 3x23y28xy+18x44y+15=03x^2 - 3y^2 - 8xy + 18x - 44y + 15 = 0

Step 4: Compare the derived equation with the given equation.

We are given the equation: x2y2+2hxy+2gx+2fy+c=0x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0 Multiply this equation by 3 to match the coefficient of x2x^2 in our derived equation: 3x23y2+6hxy+6gx+6fy+3c=03x^2 - 3y^2 + 6hxy + 6gx + 6fy + 3c = 0

Now, compare the coefficients: 6h=8    h=436h = -8 \implies h = -\frac{4}{3} 6g=18    g=36g = 18 \implies g = 3 6f=44    f=2236f = -44 \implies f = -\frac{22}{3} 3c=15    c=53c = 15 \implies c = 5

Step 5: Calculate the value of g + c + h - f.

g+c+hf=3+543(223)g + c + h - f = 3 + 5 - \frac{4}{3} - \left(-\frac{22}{3}\right) =843+223= 8 - \frac{4}{3} + \frac{22}{3} =8+183= 8 + \frac{18}{3} =8+6=14= 8 + 6 = 14

Step 6: Note the error in the original solution.

The original solution seems to have an error. The correct value for g+c+hfg + c + h - f is 14, but it incorrectly jumps to 29.

Common Mistakes & Tips

  • Sign Errors: Be very careful with signs when expanding and comparing coefficients. A small sign error can lead to a completely wrong answer.
  • Normalization: Make sure to normalize the equations by dividing or multiplying by a constant to match the coefficients before comparing. This avoids errors due to scaling.
  • Check the Arithmetic: Double-check all arithmetic calculations, especially when dealing with fractions.

Summary

We found the equations of the angle bisectors of the given lines, combined them into a single equation representing the pair of straight lines, and then compared the coefficients with the given general equation to find the values of gg, cc, hh, and ff. Finally, we calculated g+c+hfg + c + h - f to be 14.

Final Answer

The final answer is \boxed{14}, which corresponds to option (B).

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