MCQ
Which point on $x$-axis is equidistant from the points $A(7, 6)$ and $B(-3, 4)?$
- A$(0, 4)$
- B$(-4, 0)$
- ✓$(3, 0)$
- D$(0, 3)$
✓
Answer
Correct option: C.
$(3, 0)$
Let the coordinates of the point be $P(x, 0)$
Since the point lies on the $x$-axis. $P$ is equidistant from $A(7, 6)$ and $B(-3, 4)$
Using the distance from, we get
$ \mathrm{AP}^2=\mathrm{BP} $
$ \Rightarrow(7-\mathrm{x})^2+(6-0)^2=(-3-\mathrm{x})^2+(4-0)^2 $
$ \Rightarrow 49-14 \mathrm{x}+\mathrm{x}^2+36=9+6 \mathrm{x}+\mathrm{x}^2+16 $
$ \Rightarrow 85-14 \mathrm{x}=6 \mathrm{x}+25 $
$ \Rightarrow 20 \mathrm{x}=60 $
$ \Rightarrow \mathrm{x}=3$
Thus, the point $(3, 0).$
Since the point lies on the $x$-axis. $P$ is equidistant from $A(7, 6)$ and $B(-3, 4)$
Using the distance from, we get
$ \mathrm{AP}^2=\mathrm{BP} $
$ \Rightarrow(7-\mathrm{x})^2+(6-0)^2=(-3-\mathrm{x})^2+(4-0)^2 $
$ \Rightarrow 49-14 \mathrm{x}+\mathrm{x}^2+36=9+6 \mathrm{x}+\mathrm{x}^2+16 $
$ \Rightarrow 85-14 \mathrm{x}=6 \mathrm{x}+25 $
$ \Rightarrow 20 \mathrm{x}=60 $
$ \Rightarrow \mathrm{x}=3$
Thus, the point $(3, 0).$
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