A) (5/2,0)
B) (5,0)
C) (4, 0)
D) (5/4,0)
Correct Answer: D
Solution :
| Let A (x, 0) be any point on X-axis, which is equidistant from points B (1, 3) and (-1, 2). |
| AB=AC |
| \[\Rightarrow A{{B}^{2}}=A{{C}^{2}}\] |
| [squaring both sides] |
| \[\Rightarrow \,{{\left( 1-x \right)}^{2}}+{{\left( 3-0 \right)}^{2}}=\left( -1-{{x}^{2}} \right)+{{\left( 2-0 \right)}^{2}}\] |
| \[\left[ \because \,dis\tan ce\,=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \right]\] |
| \[\Rightarrow \,\,\,1+{{x}^{2}}-2x+9=1+{{x}^{2}}+2x+4\] |
| \[\Rightarrow \,\,9-4=2x+2x\] |
| \[\Rightarrow \,\,5=4\,x\Rightarrow x=5/4\] |
| Hence, point on X-axis is (5/4,0). |
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