Question 2

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Jambmaths question: 

Find the value of ${{\alpha }^{2}}+{{\beta }^{2}}$ if $\alpha +\beta =2$and the distance between the points $(1,\alpha \text{ )and (}\beta ,1)$is 3 units

Option A: 

14

Option B: 

3

Option C: 

5

Option D: 

11

Jamb Maths Solution: 

$\begin{align}  & ({{x}_{1}},{{y}_{1}})=(1,\alpha ) \\ & ({{x}_{2}},{{y}_{2}})=(\beta ,1) \\ & \text{Distance between two point is given as} \\ & d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ & d=\sqrt{{{(1-\beta )}^{2}}+{{(\alpha -1)}^{2}}} \\ & d=\sqrt{1-2\beta +{{\beta }^{2}}+{{\alpha }^{2}}-2\alpha +1} \\ & 3=\sqrt{2-2(\alpha +\beta )+{{\alpha }^{2}}+{{\beta }^{2}}} \\ & \text{Square both side} \\ & 9=2-2(2)+{{\alpha }^{2}}+{{\beta }^{2}} \\ & {{\alpha }^{2}}+{{\beta }^{2}}=9+2=11 \\\end{align}$

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