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, $AC = BC$ Or, $AC^2$
$= BC^{2 }x^2 - 10x + 106 = x^2 + 8x + 52 $
$\Rightarrow -10x + 106 = 8x + 52 $
$\Rightarrow -10x - 8x = 52 - 106 $
$\Rightarrow -18x = -54$
$\Rightarrow\ \text{x}=\frac{54}{18}$ $\Rightarrow x = 3$
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, $AC = BC$ Or, $AC^2$
$= BC^{2 }x^2 - 10x + 106 = x^2 + 8x + 52 $
$\Rightarrow -10x + 106 = 8x + 52 $
$\Rightarrow -10x - 8x = 52 - 106 $
$\Rightarrow -18x = -54$
$\Rightarrow\ \text{x}=\frac{54}{18}$ $\Rightarrow x = 3$
Hence the points on x-axis is $(3, 0)$.
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