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: Option A
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 = (frac{x_2 + x_1}{2}, frac{y_2 + y_1}{2})
= (frac{2 + 0}{2}, frac{1 + 0}{2})
= ((1, frac{1}{2}))
Gradient of QR = (frac{y_{2} – y_{1}}{x_{2} – x_{1}})
= (frac{1 – 0}{2 – 0})
= (frac{1}{2})
Gradient of PM = (frac{-1}{frac{1}{2}})
= -2
Equation of PM = y – y(_1) = m(x-x_(_1))
i.e y – 1/2 = -2(x-1)
2y – 1 = -4(x-1)
2y – 1 = -4x + 4
2y + 4x = 5