The objective is to locate a specific point situated on the X-axis. This point must satisfy the condition of being equidistant from two given points, namely A(2, -3) and B(-2, 1).
A fundamental property of any point lying on the X-axis is that its y-coordinate is always zero. Consequently, we can represent the unknown point as P(x, 0).
The core requirement is that the distance from P to A is equal to the distance from P to B. This is expressed as $PA = PB$. Squaring both sides simplifies the calculation, giving us $PA^2 = PB^2$.
The distance formula calculates the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ as $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Calculating the squared distance $PA^2$:
$PA^2 = (x - 2)^2 + (0 - (-3))^2$ $PA^2 = (x - 2)^2 + (3)^2$ $PA^2 = (x - 2)^2 + 9$Calculating the squared distance $PB^2$:
$PB^2 = (x - (-2))^2 + (0 - 1)^2$ $PB^2 = (x + 2)^2 + (-1)^2$ $PB^2 = (x + 2)^2 + 1$Equating the squared distances $PA^2$ and $PB^2$:
$(x - 2)^2 + 9 = (x + 2)^2 + 1$Expanding the squared terms:
$(x^2 - 4x + 4) + 9 = (x^2 + 4x + 4) + 1$Simplifying both sides:
$x^2 - 4x + 13 = x^2 + 4x + 5$Subtract $x^2$ from both sides to eliminate it:
$-4x + 13 = 4x + 5$Rearranging the terms to isolate x:
$13 - 5 = 4x + 4x$ $8 = 8x$Solving for x:
$x = \frac{8}{8}$ $x = 1$We determined that $x = 1$. Since the point P lies on the X-axis, its coordinates are P(x, 0). Substituting the value of x, we get P(1, 0).
Thus, the point P(1, 0) is equidistant from points A(2, -3) and B(-2, 1).
Find the co-ordinates of the centroid of a triangle whose vertices are A(1, 4), B(7, 8) and C(10, 12).