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
[tex]\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2-} \mathrm{y}_{1}\right)^{2}}[/tex]
Where [tex]x_{1}, x_{2}, y_{1}, y_{2}[/tex] are the x and y co-ordinates
The distance between A (-6, 0) and P(x, y) is:
Using the distance formulae,
[tex]\mathrm{D}=\sqrt{(\mathrm{x}+6)^{2}+(\mathrm{y}-0)^{2}}[/tex]
On taking square root we get,
[tex]D^{2}=(\mathrm{x}+6)^{2}+(\mathrm{y}-0)^{2}[/tex] --- eqn 1
The distance between B(-3, 1) and P(x, y) is:
[tex]\mathrm{D}=\sqrt{(\mathrm{x}+3)^{2}+(\mathrm{y}-1)^{2}}[/tex]
On taking square root we get
[tex]D^{2}=(\mathrm{x}+3)^{2}+(\mathrm{y}-1)^{2}[/tex] --- eqn 2
The distance between C(x, y) and P (0, 0) is:
[tex]\mathrm{D}=\sqrt{(\mathrm{x}-0)^{2}+(\mathrm{y}-0)^{2}}[/tex]
[tex]D^{2}=(\mathrm{x}-0)^{2}+(\mathrm{y}-0)^{2}[/tex] -- eqn 3
By equating equation 1 = equation 2 to find the value of y
[tex](x+6)^{2}+(y-0)^{2}=(x-0)^{2}+(y-0)^{2}[/tex]
In both the expression [tex](\mathrm{y}-0)^{2}[/tex] is common so we can cancel it.
[tex](\mathrm{x}+6)^{2}=(\mathrm{x}-0)^{2}[/tex]
On expanding we get,
[tex]x^{2}+36+12 \mathrm{x}=x^{2}[/tex]
12x = -36
x = -3.
Now find the value of y using equation 3.
[tex]\begin{array}{l}{x^{2}+y^{2}=0} \\ {(-3)^{2}+y^{2}=0} \\ {3^{2}=-y^{2}} \\ {y=-3}\end{array}[/tex]
Hence the required points are (-3,-3).
