Substitute the coordinates of P(x, 0), A(5, -3), and B(4, 2) into the distance formula and equate PA² and PB² to eliminate the square root. This gives us (x-5)² + (0-(-3))² = (x-4)² + (0-2)².

Expand and simplify the equation. The x² terms cancel out. Collect like terms to solve for x. This leads to x = 7.

The point on x-axis which is equidistant from the points (5, -3) and (4, 2) is

Answer: $(7, 0)$

Step-by-step solution

Step 1: Define the point on the x-axis

A point on the $x$ -axis always has its $y$ -coordinate equal to 0. Let this point be $P(x, 0)$ . We are given two other points, $A(5, -3)$ and $B(4, 2)$ .

Step 2: Apply the equidistant condition

The problem states that point $P$ is equidistant from points $A$ and $B$ . This means the distance from $P$ to $A$ is equal to the distance from $P$ to $B$ .

Step 3: Use the distance formula

We use the distance formula to express $PA$ and $PB$ . The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

Step 4: Set up the equation

Substitute the coordinates of $P(x, 0)$ , $A(5, -3)$ , and $B(4, 2)$ into the distance formula and equate $PA^2$ and $PB^2$ to eliminate the square root. This gives us $(x-5)^2 + (0-(-3))^2 = (x-4)^2 + (0-2)^2$ .

Step 5: Solve for x

Expand and simplify the equation. The $x^2$ terms cancel out. Collect like terms to solve for $x$ . This leads to $x = 7$ .

Step 6: State the point

Since we found $x = 7$ , the point on the $x$ -axis equidistant from the given points is $(7, 0)$ .

More from this chapter

All Coordinate Geometry solutions →

Solve your own maths question free →