Please answer the question and the answers should be in English

To find the coordinates of a point equidistant from three given points, we can set up equations using the distance formula. Let the unknown point be $P(x, y)$ and the given points be $A(2, 3)$, $B(4, 5)$, and $C(6, 1)$. The condition is $PA = PB = PC$, which implies $PA^2 = PB^2 = PC^2$.

Step 1: Set up the equation $PA^2 = PB^2$. Using the distance formula squared, $(x_2 - x_1)^2 + (y_2 - y_1)^2$: $(x - 2)^2 + (y - 3)^2 = (x - 4)^2 + (y - 5)^2$

Step 2: Expand and simplify the equation from Step 1. $x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 10y + 25$ Subtract $x^2$ and $y^2$ from both sides: $-4x - 6y + 13 = -8x - 10y + 41$ Rearrange the terms to form a linear equation: $-4x + 8x - 6y + 10y = 41 - 13$ $4x + 4y = 28$ Divide by 4: $x + y = 7 \quad (Equation 1)$

Step 3: Set up the equation $PB^2 = PC^2$. $(x - 4)^2 + (y - 5)^2 = (x - 6)^2 + (y - 1)^2$

Step 4: Expand and simplify the equation from Step 3. $x^2 - 8x + 16 + y^2 - 10y + 25 = x^2 - 12x + 36 + y^2 - 2y + 1$ Subtract $x^2$ and $y^2$ from both sides: $-8x - 10y + 41 = -12x - 2y + 37$ Rearrange the terms: $-8x + 12x - 10y + 2y = 37 - 41$ $4x - 8y = -4$ Divide by 4: $x - 2y = -1 \quad (Equation 2)$

Step 5: Solve the system of linear equations (Equation 1 and Equation 2). We have:

1. $x + y = 7$
2. $x - 2y = -1$

Subtract Equation 2 from Equation 1: $(x + y) - (x - 2y) = 7 - (-1)$ $x + y - x + 2y = 8$ $3y = 8$ $y = \frac{8}{3}$ Substitute the value of $y$ into Equation 1: $x + \frac{8}{3} = 7$ $x = 7 - \frac{8}{3}$ $x = \frac{21}{3} - \frac{8}{3}$ $x = \frac{13}{3}$

Step 6: State the coordinates of the point. The coordinates of the point equidistant from the three given points are $\left(\frac{13}{3}, \frac{8}{3}\right)$.

The coordinates of the point are $\boxed{\left(\frac{13}{3}, \frac{8}{3}\right)}$.

Send me the next one 📸