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3z1=5z2−2z3 Implies Collinear Points | Complex Numbers

JEE Maths question with a full step-by-step solution.

Question
If the complex numbers z1,z2,z3z_1, z_2, z_3 satisfy 3z1=5z2−2z33z_1 = 5z_2 - 2z_3, then z1,z2,z3z_1, z_2, z_3 lie in a/on
Acircle
Bparabola
Clinecorrect
Dhyperbola
Solution
Step 1: Rearrange the relation into a linear combination equal to zero:
3z1=5z2−2z3  ⟹  3z1−5z2+2z3=03z_1 = 5z_2 - 2z_3 \implies 3z_1 - 5z_2 + 2z_3 = 0
 Step 2: Check the sum of the (real) coefficients:
3+(−5)+2=03 + (-5) + 2 = 0
 Step 3: Use the collinearity criterion. If real numbers p1,p2,p3p_1, p_2, p_3, not all zero, satisfy p1+p2+p3=0p_1 + p_2 + p_3 = 0 and p1z1+p2z2+p3z3=0p_1 z_1 + p_2 z_2 + p_3 z_3 = 0, then the points z1,z2,z3z_1, z_2, z_3 are collinear. (Reason: p1z1+p2z2+p3z3=0p_1 z_1 + p_2 z_2 + p_3 z_3 = 0 with p1=−(p2+p3)p_1 = -(p_2 + p_3) gives p2(z2−z1)+p3(z3−z1)=0p_2(z_2 - z_1) + p_3(z_3 - z_1) = 0, so the vectors z2−z1z_2 - z_1 and z3−z1z_3 - z_1 are real scalar multiples of each other.) Step 4: Equivalently, solving for z1z_1:
z1=5z2−2z33=5z2−2z35−2z_1 = \frac{5z_2 - 2z_3}{3} = \frac{5z_2 - 2z_3}{5 - 2}
 This is the section formula: z1z_1 divides the line joining z2z_2 and z3z_3 externally in the ratio 5:25 : 2. A point dividing a segment lies on the same line. Therefore z1,z2,z3z_1, z_2, z_3 are collinear — they lie on a straight line. Correct answer: (3)
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