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# Find the point on the x-axis which is equidistant from the points (– 2, 5) and (2, – 3).

Last updated date: 15th Jun 2024
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Hint: We will let the point on the x-axis be (x, 0) as the point is equidistant from (– 2, 5) and (2, – 3). So, we will calculate the distance of the point (x, 0) to each of the two points and then compare them to find the value of x. Once we get that, we have the point on the x-axis.

We are asked to find the point on the x-axis which is equidistant from the point (– 2, 5) and (2, – 3). We know that the point on the x-axis has its y – coordinate as 0. So, let (x, 0) be the point on the x-axis which is equidistant from (– 2, 5) and (2, – 3).
The distance between the two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by the distance formula $D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ as we are given that (x, 0) is equidistant from (– 2, 5) and (2, – 3).
Now we will find their distance and then compare them to find our value of x.

Let $\left( {{x}_{1}},{{y}_{1}} \right)=\left( x,0 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( -2,5 \right)$
So, the distance between them will be given as,
${{D}_{1}}=\sqrt{{{\left( -2-x \right)}^{2}}+{{\left( 5-0 \right)}^{2}}}$
We can write ${{\left( -2-x \right)}^{2}}$ as
${{\left( -2-x \right)}^{2}}={{\left[ \left( - \right)\left( 2+x \right) \right]}^{2}}={{\left( 2+x \right)}^{2}}$
So simplifying further, we get,
$\Rightarrow {{D}_{1}}=\sqrt{{{\left( 2+x \right)}^{2}}+{{5}^{2}}}$
Similarly, we can write, $\left( {{x}_{1}},{{y}_{1}} \right)=\left( x,0 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 2,-3 \right).$
So, the distance between them will be given as,
${{D}_{2}}=\sqrt{{{\left( 2-x \right)}^{2}}+{{\left( -3-0 \right)}^{2}}}$
So simplifying further, we get,
$\Rightarrow {{D}_{2}}=\sqrt{{{\left( 2-x \right)}^{2}}+{{3}^{2}}}$
As (x, 0) is equidistant from (– 2, 5) and (2, – 3), that means ${{D}_{1}}={{D}_{2}}.$
So, we get,
$\sqrt{{{\left( 2+x \right)}^{2}}+{{5}^{2}}}=\sqrt{{{\left( 2-x \right)}^{2}}+{{3}^{2}}}$
Squaring both the sides, we get,
${{\left( 2+x \right)}^{2}}+{{5}^{2}}={{\left( 2-x \right)}^{2}}+{{3}^{2}}$
Opening the square brackets to simplify, we get,
$\Rightarrow {{x}^{2}}+4+4x+25={{x}^{2}}+4-4x+9$
Cancelling the like terms, we get,
$\Rightarrow 8x=9-25$
$\Rightarrow 8x=-16$
Dividing both the sides by 8, we get,
$\Rightarrow \dfrac{8x}{8}=\dfrac{-16}{8}$
$\Rightarrow x=-2$
So, we get, x = – 2 which means that the point on the x-axis that is equidistant from (– 2, 5) and (2, – 3) is (– 2, 0).
Hence, the required answer is (– 2, 0).

Note: We can cross-check that our solution is correct or not by the following steps. We will find the distance between (– 2, 0) and the other two points and see if they are equal or not.
(i) Distance between (– 2, 0) and (– 2, 5).
Let $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -2,0 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( -2,5 \right)$
${{D}_{1}}=\sqrt{{{\left( -2-\left( -2 \right) \right)}^{2}}+{{\left( 5-0 \right)}^{2}}}$
$\Rightarrow {{D}_{1}}=\sqrt{0+{{5}^{2}}}$
$\Rightarrow {{D}_{1}}=5$
We get the distance between (– 2, 0) and (– 2, 5) as 5 units.
(ii) Distance between (– 2, 0) and (2, – 3).
Let $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -2,0 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 2,-3 \right)$
${{D}_{2}}=\sqrt{{{\left( 2-\left( -2 \right) \right)}^{2}}+{{\left( -3-0 \right)}^{2}}}$
$\Rightarrow {{D}_{2}}=\sqrt{{{4}^{2}}+{{3}^{2}}}$
$\Rightarrow {{D}_{2}}=\sqrt{16+9}$
$\Rightarrow {{D}_{2}}=\sqrt{25}$
$\Rightarrow {{D}_{2}}=5$
So again, we get the distance between (– 2, 0) and (– 2, 5) as 5 units.