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MATHEMATICS

JAMB 2003 - Question 24

Mathematics 2003 JAMB Past Questions - Question 24: Find the equation of the locus of a point P(x,y) which is equdistant from Q(0,0) and R(2,1)

Choose the most appropriate option for the gap .

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

A: 4x+2y=5

B: 4x-2y=5

C: 2x+2y=5

D: 2x+y=5

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Correct Answer

Explanation

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 RMid point QR = x2+x1/2, y2+y1/2= 2+0/2, 1+0/2= (1,1/2) Gradient of QR = y2−y1/x2−x1= 1−0/2−0= 1/2Gradient of PM = 1/1/2= -2Equation of PM = y - y1 = m(x-x_1)i.e y - 1/2 = -2(x-1)2y - 1 = -4(x-1)2y - 1 = -4x + 42y + 4x = 5