Problem: Consider all line segments of length 4 with one end-point on the line y=x and the other end-point on the line y=2x. Find the equation of the locus of the midpoints of these line segments.
Solution
Solution: Let the end-points of the segment be A(a,a) on y=x and B(b,2b) on y=2x.
The length of AB is 4: AB=(b−a)2+(2b−a)2=4 So, (b−a)2+(2b−a)2=16 Expanding: (b−a)2+(2b−a)2=(b−a)2+(4b2−4ab+a2)=(b−a)2+4b2−4ab+a2 But (2b−a)2=4b2−4ab+a2, so: (b−a)2+(2b−a)2=(b2−2ab+a2)+(4b2−4ab+a2)=5b2−6ab+2a2=16 Let the midpoint M of AB be (2a+b,2a+2b). Let x=2a+b and y=2a+2b.
Then: a=2x−b y=2a+2b=2(2x−b)+2b=22x+b So, 2y=2x+b⟹b=2y−2x Substitute b into a: a=2x−(2y−2x)=2x−2y+2x=4x−2y Now, substitute a and b in the length equation: 5b2−6ab+2a2=16 Plug in a=4x−2y, b=2y−2x:
First, compute b2: b2=(2y−2x)2=4(y−x)2 a2: a2=(4x−2y)2=16x2−16xy+4y2 ab: ab=(4x−2y)(2y−2x)=4x(2y−2x)−2y(2y−2x)=8xy−8x2−4y2+4xy=(8xy+4xy)−8x2−4y2=12xy−8x2−4y2 But let's expand directly: ab=(4x−2y)(2y−2x)=4x⋅2y−4x⋅2x−2y⋅2y+2y⋅2x=8xy−8x2−4y2+4xy=(8xy+4xy)−8x2−4y2=12xy−8x2−4y2 Now, plug into the equation: 5b2−6ab+2a2=5⋅4(y−x)2−6(12xy−8x2−4y2)+2(16x2−16xy+4y2) Compute each term: 5b2=20(y−x)2 −6ab=−6(12xy−8x2−4y2)=−72xy+48x2+24y2 2a2=32x2−32xy+8y2
Sum: 20(y−x)2+48x2+24y2−72xy+32x2−32xy+8y2=16 But 20(y−x)2=20(y2−2xy+x2)=20y2−40xy+20x2 So sum all x2 terms: 20x2+48x2+32x2=100x2 y2 terms: 20y2+24y2+8y2=52y2 −xy terms: −40xy−72xy−32xy=−144xy
So the equation is: 100x2+52y2−144xy=16 Or, dividing both sides by 4: 25x2+13y2−36xy=4 Thus, the equation of the locus is: 25x2+13y2−36xy=4
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Source: MathNet,
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