Question
Which point on x-axis is equidistant from $(5, 9)$ and $(-4, 6)?$
✓
Answer
Let $A(5, 9)$ and $B(-4, 6)$ be the given points.
Let $C(x, 0)$ be the point on x-axis
Now,
$\text{AC}=\sqrt{(\text{x}-5)^2+(0-9)^2}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2+25-10\text{x}+(-9)^2}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2-10\text{x}+25+81}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2-10\text{x}+106}$
$\text{BC}=\sqrt{(\text{x}+4)^2+(0-6)^2}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+16+8\text{x}+(-6)^2}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+8\text{x}+16+36}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+8\text{x}+52}$
Since, $AC = BC$
$ \text { Or, } A C^2=B C^2 $
$ x^2-10 x+106=x^2+8 x+52 $
$ \Rightarrow-10 x+106=8 x+52 $
$ \Rightarrow-10 x-8 x=52-106 $
$ \Rightarrow-18 x=-54$
$\Rightarrow\ \text{x}=\frac{54}{18}$
$\Rightarrow x = 3$
Hence the points on x-axis is $(3, 0).$
Let $C(x, 0)$ be the point on x-axis
Now,
$\text{AC}=\sqrt{(\text{x}-5)^2+(0-9)^2}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2+25-10\text{x}+(-9)^2}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2-10\text{x}+25+81}$
$\Rightarrow\ \text{AC}=\sqrt{\text{x}^2-10\text{x}+106}$
$\text{BC}=\sqrt{(\text{x}+4)^2+(0-6)^2}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+16+8\text{x}+(-6)^2}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+8\text{x}+16+36}$
$\Rightarrow\ \text{BC}=\sqrt{\text{x}^2+8\text{x}+52}$
Since, $AC = BC$
$ \text { Or, } A C^2=B C^2 $
$ x^2-10 x+106=x^2+8 x+52 $
$ \Rightarrow-10 x+106=8 x+52 $
$ \Rightarrow-10 x-8 x=52-106 $
$ \Rightarrow-18 x=-54$
$\Rightarrow\ \text{x}=\frac{54}{18}$
$\Rightarrow x = 3$
Hence the points on x-axis is $(3, 0).$
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