Find the equation of the locus of a point P x y... - JAMB Mathematics 2003 Question
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 option: a
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
= 1/2
= -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
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
Please share this, thanks:
Add your answer
No responses