MCQ
The point on the $x-$ axis which is equidistant from the points $(2, -5)$ and $(-2, 9)$ is
- A$(0, – 7)$
- B$(0, 7)$
- C$(7, 0)$
- ✓$(-7, 0)$
✓
Answer
Correct option: D.
$(-7, 0)$
Let $A(2, - 5)$ and $B(-2, 9)$
Since the point is on $x-$ axis $C (x, 0)$
$\therefore\text{AC}^2=\text{BC}^2$
$\Rightarrow {(2-x)}^2+(-5-0)^2$
$=(-2-\text{x})^2+(9-0)^2$
$\Rightarrow4 + \text{x}^2 - 4\text{x} + 25 $
$= 4 + \text{x}^2 + 4\text{x} + 81$
$\Rightarrow-8\text{x}=56$
$\Rightarrow\text{x}= -7$
Therefore, the point on $x-$ axis is $(-7, 0)$
Since the point is on $x-$ axis $C (x, 0)$
$\therefore\text{AC}^2=\text{BC}^2$
$\Rightarrow {(2-x)}^2+(-5-0)^2$
$=(-2-\text{x})^2+(9-0)^2$
$\Rightarrow4 + \text{x}^2 - 4\text{x} + 25 $
$= 4 + \text{x}^2 + 4\text{x} + 81$
$\Rightarrow-8\text{x}=56$
$\Rightarrow\text{x}= -7$
Therefore, the point on $x-$ axis is $(-7, 0)$
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