A) 6
B) -11
C) 5
D) 11
Correct Answer: D
Solution :
| Let points are \[A\left( 2,\,y \right)\]and \[B\left( -4,\,3 \right)\]. |
| Here, \[\left( {{x}_{1}},\,{{y}_{1}} \right)=\left( 2,\,y \right)\] and \[\left( {{x}_{2}},\,{{y}_{2}} \right)=\left( -4,\,3 \right)\] |
| \[\therefore \] Distance between two points, |
| \[AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\] |
| [by distance formula] |
| \[\Rightarrow \,\,AB=\sqrt{{{\left( 2+4 \right)}^{2}}+{{\left( y-3 \right)}^{2}}}\] |
| \[\Rightarrow \,\,10=\sqrt{{{\left( 6 \right)}^{2}}+{{y}^{2}}+9-6y}\] |
| \[\left[ \because \,\,AB=10\,and\,{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right]\] |
| On squaring both sides, we get |
| \[\Rightarrow \,\,\,{{\left( 10 \right)}^{2}}={{\left( 6 \right)}^{2}}+{{y}^{2}}+9-6y\] |
| \[\Rightarrow \,\,100=36+{{y}^{2}}+9-6y\] |
| \[\Rightarrow \,\,\,100=45+{{y}^{2}}-6y\] |
| \[\Rightarrow \,\,\,{{y}^{2}}-6y-55=0\] |
| \[\Rightarrow \,\,\,{{y}^{2}}-11y+5y-55=0\] |
| [by factorisation] |
| \[\Rightarrow \,\,\,y\left( y-11 \right)+5\left( y-11 \right)=0\] |
| \[\Rightarrow \,\,\,\,\left( y-11 \right)\left( y+5 \right)=0\] |
| \[\Rightarrow \,\,\,y-11=0\] or \[y+5=0\] |
| \[\Rightarrow \,\,\,y=11\] or \[y=-5\] |
| Hence, the required values of y are 11 and -5. |
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