Question

The coordinates of the point which is equidistant from the points from the vertices of triangle formed by points o(0,0) A(a,0) B(0,b) is

(a) (a,b)
(b) (b,a)
(c) (a/2,b/2)
(d) (b/2,a/3)

Answer 1

Answer:

(c) (a/2,b/2)

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given:

The vertices triangle are $O(0,0), A(a,0)$ and $B(0,6).$

To find:

The coordinates of the point which is equidistant from the points from the vertices of the triangle formed by points  $O(0,0), A(a,0)$ and $B(0,b).$    

Solution:

The distance $d$ between two points  $(x_{1},y_{1})$  and  $(x_{2},y_{2})$ is given by the formula,

$d=\sqrt{(x_{1}-x_{1})^2+(x_{1} -y_{2})^2 }$

The circumcentre of a triangle is the point that is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be $O(0,0), A(a,0)$ and $B(0,b).$

Let the circumcentre of the triangle be represented by the point $R(x,y).$

So we have $OR=AR=BR$

$OR=\sqrt{(-x)^2+(-y)^2}$

$AR=\sqrt{(a-x)^2+(-y)^2}$

$BR=\sqrt{(-x)^2+(b-y)^2}$

Equating the first pair of these questions we have,

                   $OR=AR$

$\sqrt{(-x)^2+(-y)^2}=\sqrt{(a-x)^2+(-y)^2}$

Squaring on both sides of the equation we have,

  $(-x)^2+(-y)^2=(a-x)^2+(-y)^2$

             $x^2+y^2=a^2+x^2-2ax+y^2$

                    $2ax=a^2$

                    ∴ $x=\frac{a}{2}$

Equating another pair of the equations we have,

                    $OR=BR$

$\sqrt{(-x)^2+(-y)^2} =\sqrt{(-x)^2+(b-y)^2}$

Squaring on both sides of the equation we have,

  $(-x)^2+(-y)^2=(-x)^2+(b-y)^2$

             $x^2+y^2=x^2+b^2+y^2-2by$

                   $2by=b^2$

                   ∴ $y=\frac{b}{2}$

Answer:

Hence the answer is an option (c) $(\frac{a}{2},\frac{b}{2} )$.