Question

Find the coordinates of the point equidistant from the points A(1,2) B(3,-4) and C(5,-6)

Answer 1

Answer:

$\text{The coordinates of the points equidistant from the points A(1,2) B(3,-4) and C(5,-6) are }(11,2)$

Step-by-step explanation:

Given three points A(1,2) B(3,-4) and C(5,-6)

we have to find the coordinates of the point equidistant from the points.

The point that is equidistant from three points is called circumcenter which can be evaluated to find the perpendicular bisectors.

To find the perpendicular bisectors of AB:

$\text{The slope of AB=}\frac{-4-2}{3-1}=-3$

$\text{The slope of line perpendicular to AB=}\frac{1}{3}$

$\text{The mid-point of AB is=}(\frac{1+3}{2},\frac{2-4}{2})=(2,-1)$

The equation of perpendicular line is

$y=\frac{1}{3}(x-2)-1$

To find the perpendicular bisectors of AC:

$\text{The slope of AC=}\frac{-6-2}{5-1}=-2$

$\text{The slope of line perpendicular to AC=}\frac{1}{2}$

$\text{The mid-point of AC is=}(\frac{1+5}{2},\frac{2-6}{2})=(3,-2)$

The equation of perpendicular line is

$y=\frac{1}{2}(x-3)-2$

Now, solve the above two equations

$\frac{1}{3}(x-2)-1=\frac{1}{2}(x-3)-2$

$2(x-2)-6=3(x-3)-12$

$x=11$

$y=\frac{1}{2}(x-3)-2=\frac{1}{2}(11-3)-2=4-2=2$

$\text{The coordinates of the points equidistant from the points A(1,2) B(3,-4) and C(5,-6) are }(11,2)$

Answer 2

Answer:

$Required \:point = \left(3,\frac{-8}{3}\right)$

Step-by-step explanation:

Given three points A(1,2),B(3,-4) and C(5,-6).

The point which is equidistant from A,B and C is Centroid (G).

$G=\left(\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3}\right)$

$=\left(\frac{1+3+5}{3},\frac{2-4-6}{3}\right)$

$=\left(\frac{9}{3},\frac{-8}{3}\right)$

$=\left(3,\frac{-8}{3}\right)$

Therefore,

$Required \:point = \left(3,\frac{-8}{3}\right)$

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