Find the points on the x–axis which are at a distance of 2√5 from the point (7, –4). How many such points are there

Solution:

Given, the points on the x-axis are at a distance of 2√5 from the point (7, -4).

We have to find the number of points that are at a distance of 2√5 from the point (7, -4).

The points on the x-axis mean the point represents the form (x, 0)

Let the given point be B = (7, -4)

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

The distance between two points P (x₁ , y₁) and Q (x₂ , y₂) is

√[(x₂ - x₁)² + (y₂ - y₁)²]

Distance between A(x, 0) and B(7, -4) = 2√5

√[(x₂ - x₁)² + (y₂ - y₁)²] = 2√5

√[(7 - x)² + (-4 - 0)²] = 2√5

√[(7 - x)² + (-4)²] = 2√5

On squaring both sides,

(7 - x)² + 16 = 20

(a - b)² = a² - 2ab + b²

So, (7 - x)² = (7)² - 2(7)(x) + (x)²

= 49 - 14x + x²

49 - 14x + x² + 16= 20

x² - 14x + 49 - 20 + 16 = 0

x² - 14x + 45 = 0

x² - 9x - 5x + 45 = 0

x(x - 9) - 5(x - 9) = 0

(x - 5)(x - 9) = 0

Now, x - 5 = 0

x = 5

Also, x - 9 =0

x = 9

Therefore, the points (5, 0) and (9, 0) are at a distance of 2√5 from the point (7, -4).

✦ Try This: Find the coordinates of the point R on the line segment joining the points P(- 1, 3) and Q(2, 5) such that PR = 3PQ/5.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 2

Summary:

The points on the x–axis (5, 0) and (9, 0) are at a distance of 2√5 from the point (7, –4). There are 2 such points

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