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