Find the equation of the locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1).
A.
4x + 2y = 5
B.
4x - 2y = 5
C.
2x + 2y = 5
D.
2x + y = 5
Correct Answer: 4x + 2y = 5
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 R
Mid 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
= 1/2
Gradient of PM = −1/(1/2)
= -2
Equation of PM = y - y1
= m(x-x_1)
i.e y - 1/2 = -2(x-1)
2y - 1 = -4(x-1)
2y - 1 = -4x + 4
2y + 4x = 5
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
= (1,1/2)
Gradient of QR = (y2−y1)/(x2−x1)
= 1−0/2−0
= 1/2
Gradient of PM = −1/(1/2)
= -2
Equation of PM = y - y1
= m(x-x_1)
i.e y - 1/2 = -2(x-1)
2y - 1 = -4(x-1)
2y - 1 = -4x + 4
2y + 4x = 5
