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Difficulty: MediumLoci and Geometric Constructions

Determine the equation of the locus of a point P(x,y)P(x, y) that is equidistant from the two parallel lines 2x−3y+6=02x - 3y + 6 = 0 and 2x−3y−4=02x - 3y - 4 = 0.

Answer: 2x - 3y + 1 = 0 / 2x-3y+1=0 / 2x - 3y = -1 / 2x-3y=-1

Answer

2x−3y+1=02x - 3y + 1 = 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, defined by ax+by+c1+c22=0ax + by + \frac{c_1 + c_2}{2} = 0. Substituting c1=6c_1 = 6 and c2=−4c_2 = -4 yields 6+(−4)2=1\frac{6 + (-4)}{2} = 1, giving the equation 2x−3y+1=02x - 3y + 1 = 0.

Step-by-Step Solution

1
Identify the geometric principle for the locus between two parallel lines.
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 third parallel line given by ax+by+c1+c22=0ax + by + \frac{c_1 + c_2}{2} = 0.
Points equidistant from two parallel lines lie on a parallel line midway between them.
2
Calculate the average of the constant terms c1=6c_1 = 6 and c2=−4c_2 = -4.
cmid=6+(−4)2=22=1c_{mid} = \frac{6 + (-4)}{2} = \frac{2}{2} = 1.
The midpoint constant term is the arithmetic mean of the two original constants.
3
Construct the equation of the locus line.
2x−3y+1=02x - 3y + 1 = 0.
Combining the common linear coefficients 2x−3y2x - 3y with the calculated midpoint constant 11 gives the required equation.

Key Concept

Locus equidistant from two parallel lines
Estimated Time:1m 30s
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Topics

Loci and Geometric Constructions

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