What is the point on the x-axis which is equidistant from 2 5 and 2 9?

(- 7, 0)
The point on the x-axis which is equidistant from (2, – 5) and (- 2, 9) is (- 7, 0).

What point on x-axis is equidistant from 5’4 and 2 3?

(2,0)

Which point on x-axis is equidistant from − 2 5 and 2 − 3?

So, we get, x = – 2 which means that the point on the x-axis that is equidistant from (– 2, 5) and (2, – 3) is (– 2, 0). Hence, the required answer is (– 2, 0).

What is the equation of x 0?

x = 0 is the equa Answer : Here, x = 0 is the equation of y-axis. Since, if we plot, x = 0 all the points will lie on y-axis irrespective of the value of y.

What is equidistant formula?

The distance between any two given points can be calculated with the help of the distance formula: d = √[(x2−x1)2+(y2−y1)2] √ [ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ] ; here [x1 x 1 and y1 y 1 )] are the coordinates of one point and [x2 x 2 and y2 y 2 ] are the coordinates of the other point.

What point on the x-axis is equidistant from the point A 7 6 & B (- 3 4?

Hence, point on x-axis which is equidistant from the points A(7,6) and B(-3,4) is (3,0)

How do you calculate equidistant points?

A point is said to be equidistant from two other points when it is at an equal distance away from both of them. The distance between any two given points can be calculated by using the distance formula with the help of the coordinates of the two points.

How do you find the equidistant points on the Y axis?

Let us consider the point on the y-axis to be (0, y) as the x-coordinate is 0 on the y-axis. So,the point (0, y) is equidistant from (- 5, 2) and (9, -2). Since their distance is equal we can apply the distance formula. Now simplifying step by step to find the value of y.

How do you find the coordinates of a point equidistant?

Let P (x, y) be the point that is equidistant from the given three points. According to distance formula, we have distance between any two points (x, y) and (x1 , y1 ) = √(x – x1 )2 + (y – y1 )2.

What is an equation for X?

Isolate “x” on one side of the algebraic equation by subtracting the sum that appears on the same side of the equation as the “x.” For example, in the equation “x + 5 = 12”, rewrite the equation as “x = 12 – 5” and solve for “x.” The solution is “x = 7.”