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Find the equation of the locus of a point A(x, y) which is equidistant from...

Mathematics
JAMB 2019

Find the equation of the locus of a point A(x, y) which is equidistant from B(0, 2) and C(2, 1)

  • A. 4x + 2y = 3
  • B. 4x - 3y = 1
  • C. 4x - 2y = 1
  • D. 4x + 2y = -1
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Correct Answer: Option C
Explanation

Since A(x, y) is the point of equidistance between B and C, then 

AB = AC

(AB)\(^2\) = (AC)\(^2\)

Using the distance formula, 

(x - 0)\(^2\) + (y - 2)\(^2\) = (x - 2)\(^2\) + (y - 1)\(^2\)

x\(^2\) + y\(^2\) - 4y + 4 = x\(^2\) - 4x + 4 + y\(^2\) - 2y + 1

x\(^2\) - x\(^2\) + y\(^2\) - y\(^2\) + 4x - 4y + 2y = 5 - 4

4x - 2y = 1

There is an explanation video available below.


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Explanation Video

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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995