Question

find the coordinate of the point which is equidistant from three given point A(5 1) B(3 7) C (7 1)​

Answer 1

Given: Points -

• A(5, 1)
• B(3, 7)
• C(7, 1)

To find: The coordinate of the point equidistant from those points.

Answer:

Let's assume that the coordinate is D(x, y).

Hence, DA = DB = DC.

Let's find the distance of DA first.

Formula to find the distance between two points:

$\sf Distance\ =\ \sqrt{\Bigg(x_2\ -\ x_1\Bigg)^2\ +\ \Bigg(y_2\ -\ y_1\Bigg)^2}$

From points A(5, 1) and D(x, y), we have:

• x₁ = 5
• x₂ = x
• y₁ = 1
• y₂ = y

Substituting them into the formula,

$\sf Distance\ =\ \sqrt{\Bigg(x\ -\ 5\Bigg)^2\ +\ \Bigg(y\ -\ 1\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(x^2\ -\ 10x\ +\ 25\Bigg)\ +\ \Bigg(y^2\ -\ 2y\ +\ 1\Bigg)}\\\\\\\\Distance\ =\ \sqrt{x^2\ -\ 10x\ +\ 25\ +\ y^2\ -\ 2y\ +\ 1}\\\\\\\\\bf Distance\ =\ \sqrt{x^2\ -\ 10x\ -\ 2y\ +\ 26\ +\ y^2}$

Now, let's find the distance of DB.

From points B(3, 7) and D(x, y), we have:

• x₁ = 3
• x₂ = x
• y₁ = 7
• y₂ = y

Substituting them into the formula,

$\sf Distance\ =\ \sqrt{\Bigg(x\ -\ 3\Bigg)^2\ +\ \Bigg(y\ -\ 7\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(x^2\ -\ 6x\ +\ 9\Bigg)\ +\ \Bigg(y^2\ -\ 14y\ +\ 49\Bigg)}\\\\\\\\Distance\ =\ \sqrt{x^2\ -\ 6x\ +\ 9\ +\ y^2\ -\ 14y\ +\ 49}\\\\\\\\\bf Distance\ =\ \sqrt{x^2\ -\ 6x\ -\ 14y\ +\ 58\ +\ y^2}$

Now, let's find the distance of DC.

From points C(7, 1) and D(x, y), we have:

• x₁ = 7
• x₂ = x
• y₁ = 1
• y₂ = y

Substituting them into the formula,

$\sf Distance\ =\ \sqrt{\Bigg(x\ -\ 7\Bigg)^2\ +\ \Bigg(y\ -\ 1\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(x^2\ -\ 14x\ +\ 49\Bigg)\ +\ \Bigg(y^2\ -\ 2y\ +\ 1\Bigg)}\\\\\\\\Distance\ =\ \sqrt{x^2\ -\ 14x\ +\ 49\ +\ y^2\ -\ 2y\ +\ 1}\\\\\\\\\bf Distance\ =\ \sqrt{x^2\ -\ 14x\ -\ 2y\ +\ 50\ +\ y^2}$

Now, let's equate all the distances to each other.

$\sf \sqrt{x^2\ -\ 10x\ -\ 2y\ +\ 26\ +\ y^2}\ =\ \sqrt{x^2\ -\ 6x\ -\ 14y\ +\ 58\ +\ y^2}\ =\\\\ \sqrt{x^2\ -\ 14x\ -\ 2y\ +\ 50\ +\ y^}$

Squaring all the roots,

$\sf x^2\ -\ 10x\ -\ 2y\ +\ 26\ +\ y^2\ =\ x^2\ -\ 6x\ -\ 14y\ +\ 58\ +\ y^2\ =\ x^2\ -\ 14x\ -\ 2y\ +\ 50\ +\ y^2$

$\sf -10x\ -\ 2y\ +\ 26\ =\ -6x\ -\ 14y\ +\ 58\ =\ -\ 14x\ -\ 2y\ +\ 50$

Let's consider the first two parts.

$\sf -10x\ -\ 2y\ +\ 26\ =\ -6x\ -\ 14y\ +\ 58\\\\\\-10x\ +\ 6x\ -\ 2y\ +\ 14y\ =\ 58\ -\ 26\\\\\\-4x\ +\ 12y\ =\ 32\\\\\\-x\ +\ 3y\ =\ 8\ \longrightarrow \tt [Equation\ 1]$

Now, let's consider the first and the last parts.

$\sf -10x\ -\ 2y\ +\ 26\ =\ -14x\ -\ 2y\ +\ 50\\\\\\-10x\ +\ 14x\ -\ 2y\ +\ 2y\ =\ 50\ -\ 26\\\\\\4x\ =\ 32\\\\\\x\ =\ \dfrac{32}{4}\\\\\\\bf x\ =\ 8$

Now that we know the value of x, let's substitute it into Equation 1.

$\sf -\ (8)\ +\ 3y\ =\ 8\\\\\\-8\ +\ 3y\ =\ 8\\\\3y\ =\ 0\\\\y\ =\ \dfrac{0}{3}\\\\\\\bf y\ =\ 0$

Therefore, the coordinate of the point which is equidistant from points A(5, 1), B(3, 7) and C(7, 1) is D(8, 0).