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Exercise 6.5 · Q3

Q.Which of the following points lies on the locus of 3x2+3y2−8x−12y+17=03x^2+3y^2-8x-12y+17=0?

(1) (0,0)(0,0)
(2) (−2,3)(-2,3)
(3) (1,2)(1,2)
(4) (0,−1)(0,-1)
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✓ Free question

Substitute each candidate point directly into 3x2+3y2−8x−12y+17=03x^2+3y^2-8x-12y+17=0; only (1,2)(1,2) makes the left side zero.

A point lies on the locus exactly when its coordinates satisfy the equation, so each option is tested by direct substitution.

Step 1. Test (0,0)(0,0). 3(0)+3(0)−8(0)−12(0)+17=17≠03(0)+3(0)-8(0)-12(0)+17 = 17 \ne 0. Rejected.

Step 2. Test (−2,3)(-2,3). 3(4)+3(9)−8(−2)−12(3)+17=12+27+16−36+17=36≠03(4)+3(9)-8(-2)-12(3)+17 = 12+27+16-36+17 = 36 \ne 0. Rejected.

Step 3. Test (1,2)(1,2). 3(1)+3(4)−8(1)−12(2)+17=3+12−8−24+17=0.3(1)+3(4)-8(1)-12(2)+17 = 3+12-8-24+17 = 0. Satisfied!

Step 4. Test (0,−1)(0,-1) for completeness. 3(0)+3(1)−8(0)−12(−1)+17=3+12+17=32≠03(0)+3(1)-8(0)-12(-1)+17 = 3+12+17 = 32 \ne 0. Rejected.

Only (1,2)(1,2) satisfies the equation, so it is the point on the locus (the given equation is in fact a circle after dividing by 33: x2+y2−83x−4y+173=0x^2+y^2-\tfrac83x-4y+\tfrac{17}{3}=0).

✓Final answer

Option (3): (1,2)(1,2)

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