Jambmaths question:
The locus of a point equidistant from two point P(6,2) and R (4,2) is perpendicular bisector of PR passing through
Option A:
(4, 5)
Option B:
(5, 2)
Option C:
(1, 0)
Option D:
(0, 1)
Jamb Maths Solution:
$\begin{align} & \text{The bisector of }AB\text{ will be} \\ & (x,y)=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\ & (x,y)=\left( \frac{6+4}{2},\frac{2+2}{2} \right) \\ & (x,y)=\left( 5,2 \right) \\\end{align}$
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