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Difficulty: MediumLoci and Geometric Constructions

A point P(x,y)P(x, y) moves in the Cartesian plane such that it is at all times equidistant from two parallel lines given by the equations 3x+4y−12=03x + 4y - 12 = 0 and 3x+4y+4=03x + 4y + 4 = 0. The locus of PP intersects the straight line x−2y−8=0x - 2y - 8 = 0 at the point (a,b)(a, b). What is the value of a−ba - b?

Answer: 6

Answer

The value of a−ba - b is 6.
The locus of a point moving equidistant between two parallel lines is the parallel line lying midway between them. Combining the parallel line equations 3x+4y−12=03x + 4y - 12 = 0 and 3x+4y+4=03x + 4y + 4 = 0 yields the locus line 3x+4y−4=03x + 4y - 4 = 0. Solving the system formed by this locus line and x−2y−8=0x - 2y - 8 = 0 gives x=4x = 4 and y=−2y = -2. Therefore, a=4a = 4 and b=−2b = -2, so a−b=4−(−2)=6a - b = 4 - (-2) = 6.

Step-by-Step Solution

1
Find the equation of the locus of point P
Locus equation: 3x+4y−4=03x + 4y - 4 = 0
The locus of points equidistant from two parallel lines ax+by+c1=0ax + by + c_1 = 0 and ax+by+c2=0ax + by + c_2 = 0 is a parallel line midway between them, given by ax+by+c1+c22=0ax + by + \frac{c_1 + c_2}{2} = 0.
2
Solve the simultaneous equations to find the intersection point (a,b)(a, b)
a=4a = 4 and b=−2b = -2
Substitute x=2y+8x = 2y + 8 into 3x+4y−4=03x + 4y - 4 = 0 to get 3(2y+8)+4y−4=03(2y + 8) + 4y - 4 = 0, which yields 10y=−2010y = -20, so y=−2y = -2 and x=4x = 4.
3
Calculate the difference a−ba - b
6
Subtract b=−2b = -2 from a=4a = 4 to obtain 4−(−2)=64 - (-2) = 6.

Key Concept

Locus of points equidistant from two parallel lines and intersection of straight lines
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Topics

Loci and Geometric Constructions

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