Question

Find the point equidistant from the three points (-6,0),(-3,1) and (0,0).​

Answer 1

Answer:

The point equidistant from the three points (-6,0),(-3,1) and (0,0) is (-3,-3)

Solution:

The given three points are A (-6, 0), B (–3, 1) and C (0, 0)

Let P (x, y) be the point equidistant from these three points.

Distance between two points is given as  

$\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2-} \mathrm{y}_{1}\right)^{2}}$

Where $x_{1}, x_{2}, y_{1}, y_{2}$ are the x and y co-ordinates

The distance between A (-6, 0) and P(x, y) is:

Using the distance formulae,

$\mathrm{D}=\sqrt{(\mathrm{x}+6)^{2}+(\mathrm{y}-0)^{2}}$

On taking square root we get,

$D^{2}=(\mathrm{x}+6)^{2}+(\mathrm{y}-0)^{2}$ --- eqn 1

The distance between B(-3, 1) and P(x, y) is:

$\mathrm{D}=\sqrt{(\mathrm{x}+3)^{2}+(\mathrm{y}-1)^{2}}$

On taking square root we get

$D^{2}=(\mathrm{x}+3)^{2}+(\mathrm{y}-1)^{2}$ --- eqn 2

The distance between C(x, y) and P (0, 0) is:

$\mathrm{D}=\sqrt{(\mathrm{x}-0)^{2}+(\mathrm{y}-0)^{2}}$

$D^{2}=(\mathrm{x}-0)^{2}+(\mathrm{y}-0)^{2}$ -- eqn 3

By equating equation 1 = equation 2 to find the value of  y

$(x+6)^{2}+(y-0)^{2}=(x-0)^{2}+(y-0)^{2}$

In both the expression $(\mathrm{y}-0)^{2}$ is common so we can cancel it.

$(\mathrm{x}+6)^{2}=(\mathrm{x}-0)^{2}$

On expanding we get,

$x^{2}+36+12 \mathrm{x}=x^{2}$

12x = -36

x = -3.

Now find the value of y using equation 3.

$\begin{array}{l}{x^{2}+y^{2}=0} \\ {(-3)^{2}+y^{2}=0} \\ {3^{2}=-y^{2}} \\ {y=-3}\end{array}$

Hence the required points are (-3,-3).