Find the equation of the locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1).

Assessment: JAMB UTME - Mathematics - 2003 Subject: General Mathematics

Question 1 Report

Find the equation of the locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1).
Answer Details
Locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1) is the perpendicular bisector of the straight line joining Q and R
Mid point QR = (x2+x1)/2 . (y2+y1)/2
= 2+0/2 . 1+0/2
Gradient of Qr = y2-y1
x2-x1
= 1-0
2-0

= 1/2
Gradient of PM(M)= -1
1/2

= -2
Equation of Pm = y - y1 = m(x-x1)
i.e y - 1/2 = -2(x-1)
2y - 1 = -4(x-1)
2y - 1 = -4x + 4
2y + 4x = 5
Read lesson note on Loci (JAMB)
Read lesson note on Loci (WAEC)
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