A) \[\left( 0,\,-\frac{5}{3} \right)\]
B) \[\left( 2,\,\,1 \right)\]
C) \[\left( \frac{5}{4},\,\,0 \right)\]
D) \[\left( \frac{11}{4},\,\,2 \right)\]
Correct Answer: B
Solution :
Given, \[4x-3y=5\]and \[2{{x}^{3}}-3{{y}^{2}}=12\] \[\therefore \] \[2{{\left( \frac{5+3y}{4} \right)}^{2}}-3{{y}^{2}}=12\] \[\Rightarrow \] \[\frac{(25+9{{y}^{2}}+30y)}{8}-3{{y}^{2}}=12\] \[\Rightarrow \] \[15{{y}^{2}}-30y+71=0\] \[\Rightarrow \]\[y=\frac{30\pm \sqrt{900-4260}}{30}=1\pm \frac{\sqrt{-3360}}{30}\] Also, \[2{{x}^{2}}-3{{\left( \frac{4x-5}{3} \right)}^{2}}=12\] \[\Rightarrow \] \[10{{x}^{2}}-40x+61=0\] \[\Rightarrow \] \[x=\frac{40\pm \sqrt{1600-4\times 10\times 61}}{2\times 10}\] \[=\frac{40\pm \sqrt{-840}}{20}=2\pm \frac{\sqrt{-840}}{20}\] \[\therefore \]Points are \[A\left( 2+\frac{\sqrt{-840}}{20},\,1+\frac{\sqrt{-3360}}{30} \right)\] and \[B\left( 2-\frac{\sqrt{-840}}{1},1-\frac{\sqrt{-3360}}{30} \right).\] \[\therefore \]Midpoint of \[AB\]is \[(2,\,1).\]
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