A) (0, 3)
B) (4, 3)
C) (3, 0)
D) None of these
Correct Answer: C
Solution :
| Let \[A\left( x,\,0 \right)\]be any point on the X-axis, which is equidistant from points B (7, 6) and C(-3,4). |
| \[\therefore AB=AC\] |
| \[\Rightarrow A{{B}^{2}}=A{{C}^{2}}\] |
| [on squaring both sides] |
| \[\Rightarrow \,\,{{\left( 7-x \right)}^{2}}+{{\left( 6-0 \right)}^{2}}={{\left( -3-x \right)}^{2}}+{{\left( 4-0 \right)}^{2}}\] |
| \[\left[ \because \,\,dis\tan ce=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \right]\] |
| \[\Rightarrow \,\,49+{{x}^{2}}-14x+36=9+{{x}^{2}}+6x+16\] |
| \[\left[ \because \,\,{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right]\] |
| \[\Rightarrow \,\,\,20x=60\,\,\Rightarrow \,\,x=3\] |
| Hence, the required point is A (3,0). |
You need to login to perform this action.
You will be redirected in
3 sec
