What are the coordinates of the point on the x-axis which is equidistant from the points 7 6?

Class-X Maths Coordinate Geometry

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Question
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Given:

Points on the cartesian plane (7, 6) and (-3, 4).

Concept used:

Distance between (x1, y1) and (x2, y2) = √{(x2 - x1)2 + (y2 - y1)2}

Calculation:

Let the point on the x axis be (x, 0)

Distance between (x, 0) and (7, 6)

⇒ √{(x2 - x1)2 + (y2 - y1)2}

⇒ √{(7 - x)2 + (6 - 0)2}

⇒ √(x2 - 14x + 85)

Distance between (x, 0) and (- 3, 4)

⇒ √{(x2 - x1)2 + (y2 - y1)2}

⇒ √{(- 3 - x)2 + (4 - 0)2}

⇒ √(x2 + 6x + 25)

As the point (x, 0) is equidistant from the two points, both the distance calculate are equal

⇒ √(x2 - 14x + 85) = √(x2 + 6x + 25)

⇒ x2 - 14x + 85 = x2 + 6x + 25

⇒ 20x = 60

⇒ x = 3

Then the point is (3, 0)

∴ (3, 0) is the equidistance from the point (7, 6) and (-3, 4).

Solution:

The distance between ant two points can be measured using the Distance Formula which is given by: Distance Formula = √ [(x₂ - x₁)2 + (y₂ - y₁)2]

Let's assume a point P on the x-axis which is of the form P(x, 0).

We have to find a point on the x-axis which is equidistant from A (2, - 5) and B (- 2, 9).

To find the distance between P and A, substitute the values of P (x, 0) and A (2, - 5) in the distance formula.

PA = √(x - 2)² + (0 - (- 5))²

= √(x - 2)² + (5)² --------- (1)

To find the distance between P and B, substitute the values of P (x, 0) and B (- 2, 9) in the distance formula.

PB = √(x - (- 2))² + (0 - 9)²

= √(x + 2)² + (- 9)² ---------- (2)

By the given condition, these distances are equal in measure.

Hence, PA = PB

√(x - 2)² + (5)² = √(x + 2)² + (- 9)² [From equation (1) and (2)]

Squaring on both sides, we get

(x - 2)2 + 25 = (x + 2)2 + 81

x2 + 4 - 4x + 25 = x2 + 4 + 4x + 81

8x = 25 - 81

8x = - 56

x = - 7

Therefore, the point equidistant from the given points on the x-axis is (- 7, 0).

☛ Check: NCERT Solutions for Class 10 Maths Chapter 7

Video Solution:

NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.1 Question 7

Summary:

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

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Determine if the points (1, 5), (2, 3) and (- 2, - 11) are collinear.
Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.

The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula

`d = sqrt((x_1-x_2)^2 + (y_1 - y_2)^2)`

Here we are to find out a point on the x−axis which is equidistant from both the points A(7,6) and B(−3,4).

Let this point be denoted as C(x, y).

Since the point lies on the x-axis the value of its ordinate will be 0. Or in other words, we have y = 0.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C’

`AC = sqrt((7 - x)^2 + (6 - y)^2)`

`= sqrt((7 - x)^2 + (6 - 0)^2)`

`AC = sqrt((7-x)^2 + (6)^2)`

`BC= sqrt((-3-x)^2 + (4- y)^2)`

`= sqrt((-3-x)^2 + (4 - 0)^2)`

`BC = sqrt((-3-x)^2 + (4)^2)`