The distance of the point P (2, 3) from the x-axis is
a. 2
b. 3
c. 1
d. 5

Solution:

We know that the formula to find the distance between two points P(x₁, y₁) and Q(x₂, y₂) is

Distance=$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

It is given that

We have to find the distance of the point P (2, 3) from x-axis i.e. (2, 0)

Substituting these values in the equation we get

Distance=$\sqrt{(2-2)^{2}+(0 - 3)^{2}}$

So we get

Distance of P = √9

Distance of P = 3 units

Therefore, the distance of the point P is 3 units.

✦ Try This: The distance of the point P (4, 2) from the x-axis is

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

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NCERT Exemplar Class 10 Maths Exercise 7.1 Problem 1

The distance of the point P (2, 3) from the x-axis is a. 2, b. 3, c. 1, d. 5

Summary:

The distance of the point P (2, 3) from the x-axis is 3 units

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☛ Related Questions:

• The distance between the points A (0, 6) and B (0, –2) is a. 6 ,b. 8, c. 4, d. 2
• The distance of the point P (–6, 8) from the origin is a. 8, b. 2 √7, c. 10, d. 6
• The distance between the points (0, 5) and (–5, 0) is a. 5, b. 5 √2, c. 2 √5, d. 10