Complex Numbers and Quadratic Equations: Mathematics | JEE Main
Let z1=2+3i and z2=3+4i. The set S={z∈C:∣z−z1∣2−∣z−z2∣2=∣z1−z2∣2} represents a
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Hint 1 of 3
Let z=x+iy. What is the algebraic expression for ∣z−z1∣2−∣z−z2∣2 given z1=2+3i and z2=3+4i?
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Step-by-step solutionView
Correct answer
The equation reduces to the straight line x+y=7, whose axis intercepts are 7 and 7, giving a sum of 14.
Option analysis
Why each option works or fails
A · hyperbola with the length of the transverse axis 7
Confusing the condition ∣z−z1∣2−∣z−z2∣2=c with the definition of a hyperbola, which requires unsquared distances ∣z−z1∣−∣z−z2∣=extconstant. Recognize that the difference of squared Euclidean distances expands into linear terms in x and y because the quadratic terms x2+y2 cancel out, producing a straight line rather than a conic section.
B · hyperbola with eccentricity 2
Assuming that a difference between two distance terms involving fixed points must always define a hyperbola. Expand z as x+iy; the terms ∣z−z1∣2 and ∣z−z2∣2 both contain x2+y2, which cancel upon subtraction, leaving a degree-1 polynomial equation representing a straight line.
C · straight line with the sum of its intercepts on the coordinate axes equals −18
Making a sign error when expanding the squared terms or solving for the intercepts, leading to negative intercepts. Carefully expand (x−2)2+(y−3)2−[(x−3)2+(y−4)2]=2, yielding 2x+2y−12=2, or x+y=7, which has positive intercepts 7 and 7.
D · straight line with the sum of its intercepts on the coordinate axes equals 14
None. This is the correct option. Substituting z=x+iy and simplifying the squared distances yields x+y=7, with intercepts a=7 and b=7, summing to 14.
Reviewed route
Solution
StepWorking
01given
z1=2+3i, z2=3+4i, and the locus condition ∣z−z1∣2−∣z−z2∣2=∣z1−z2∣2.
02approach
Substitute z=x+iy and express each squared modulus in terms of real coordinates x and y. Then simplify to determine the resulting curve.
03execute
First, evaluate the right-hand side: ∣z1−z2∣2=∣(2−3)+i(3−4)∣2=∣−1−i∣2=(−1)2+(−1)2=2. Next, write the terms on the left: ∣z−z1∣2=(x−2)2+(y−3)2=x2−4x+4+y2−6y+9, and ∣z−z2∣2=(x−3)2+(y−4)2=x2−6x+9+y2−8y+16. Subtracting them gives: [(x−2)2+(y−3)2]−[(x−3)2+(y−4)2]=(−4x+6x)+(−6y+8y)+(13−25)=2x+2y−12. Setting this equal to 2: 2x+2y−12=2⟹2x+2y=14⟹x+y=7.
04execute
The line is x+y=7, or in intercept form 7x+7y=1. The x-intercept is 7 and the y-intercept is 7. The sum of the intercepts on the coordinate axes is 7+7=14.
✓verify
Since both ∣z−z1∣2 and ∣z−z2∣2 have quadratic terms x2+y2 with identical coefficients (1), they cancel out completely under subtraction. Thus, the equation must be linear (a straight line), immediately ruling out hyperbolas (options 0 and 1). At x=0,y=7 and y=0,x=7, sum of intercepts is 14.
Hints that build this answer step by step
Let z=x+iy. What is the algebraic expression for ∣z−z1∣2−∣z−z2∣2 given z1=2+3i and z2=3+4i?
Why is this a straight line and not a hyperbola, even though it involves the difference between distances to two points?
A hyperbola is defined by the difference of unsquared distances, ∣z−z1∣−∣z−z2∣=constant. When the distances are squared, the x2+y2 terms cancel out completely, yielding a first-degree equation in x and y, which is always a straight line.