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Zorluk: OrtaLoci and Geometric Constructions

A point P(x,y)P(x, y) moves in the Cartesian plane such that its distance from the fixed point A(2,3)A(2, 3) is equal to its distance from the fixed point B(4,1)B(4, 1). What is the equation of the locus of PP?

Cevap: x - y - 1 = 0 / x - y = 1 / y = x - 1 / x-y-1=0 / x-y=1 / y=x-1

Cevap

The equation of the locus of PP is x−y−1=0x - y - 1 = 0 (or y=x−1y = x - 1).
The set of all points equidistant from two fixed points A(2,3)A(2, 3) and B(4,1)B(4, 1) forms the perpendicular bisector of segment ABAB. Equating the squared distances (x−2)2+(y−3)2=(x−4)2+(y−1)2(x-2)^2 + (y-3)^2 = (x-4)^2 + (y-1)^2 and simplifying yields the linear equation x−y−1=0x - y - 1 = 0 (or y=x−1y = x - 1).

Adım Adım Çözüm

1
Set up the distance equality condition using the distance formula.
sqrt(x−2)2+(y−3)2=sqrt(x−4)2+(y−1)2\\sqrt{(x - 2)^2 + (y - 3)^2} = \\sqrt{(x - 4)^2 + (y - 1)^2}
The locus of a point P(x,y)P(x,y) equidistant from two points AA and BB satisfies PA=PBPA = PB.
2
Square both sides to eliminate the square roots and expand the terms.
(x−2)2+(y−3)2=(x−4)2+(y−1)2impliesx2−4x+4+y2−6y+9=x2−8x+16+y2−2y+1(x - 2)^2 + (y - 3)^2 = (x - 4)^2 + (y - 1)^2 \\implies x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 2y + 1
Squaring removes the radical sign, allowing algebraic simplification.
3
Subtract x2+y2x^2 + y^2 from both sides and collect linear terms.
−4x−6y+13=−8x−2y+17-4x - 6y + 13 = -8x - 2y + 17
The quadratic terms cancel out since the locus equidistant from two points is a linear equation (perpendicular bisector).
4
Rearrange all terms to one side and simplify.
(-4x + 8x) + (-6y + 2y) + (13 - 17) = 0 \\implies 4x - 4y - 4 = 0 \\implies x - y - 1 = 0
Dividing the entire linear equation by 44 gives the equation in simplest form.

Anahtar Kavram

Locus equidistant from two fixed points (Perpendicular Bisector of a line segment)

Alternatif Yöntem

Find the midpoint MM of ABAB, M=left(frac2+42,frac3+12right)=(3,2)M = \\left(\\frac{2+4}{2}, \\frac{3+1}{2}\\right) = (3, 2). Calculate the gradient of ABAB, m1=frac1−34−2=frac−22=−1m_1 = \\frac{1 - 3}{4 - 2} = \\frac{-2}{2} = -1. The perpendicular gradient is m2=−frac1−1=1m_2 = -\\frac{1}{-1} = 1. Use point-slope form: y−2=1(x−3)impliesy=x−1y - 2 = 1(x - 3) \\implies y = x - 1 or x−y−1=0x - y - 1 = 0.
Tahmini Süre:1m 30s
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Konular

Loci and Geometric Constructions

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