Question 27

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Jambmaths question: 

P(–6,1) and Q(6,6) are two ends of the diameter of a circle the radius.

  

 

Option A: 

6.5units

Option B: 

13.0units

Option C: 

3.5units 

Option D: 

7.0units

Jamb Maths Solution: 

$\begin{align}  & \text{Distance between }PQ\text{ is }d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ & ({{x}_{1}},{{y}_{1}})=(-6,1) \\ & ({{x}_{2}},{{y}_{2}})=(6,6) \\ & d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ & d=\sqrt{{{[6-(-6)]}^{2}}+{{(6-1)}^{2}}} \\ & d=\sqrt{{{12}^{2}}+{{5}^{2}}}=\sqrt{169} \\ & d=13 \\ & radius=\frac{diameter}{2}=\frac{13}{2}=6.5units \\\end{align}$

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