Question 31

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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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