The distance of the point P (–6, 8) from the origin is
a. 8
b. 2√7
c. 10
d. 6

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 (-6, 8) from origin i.e. (0, 0)

Substituting these values in the equation we get

Distance = \(\sqrt{(0-(-6))^{2}+(0 - 8)^{2}}\)

So we get

Distance of point P = √(36 + 64) = √100 = 10

Distance of point P = 10 units

Therefore, the distance of the point P (–6, 8) from the origin is 10 units.

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

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 (-4, 2) from origin i.e. (0, 0)

Substituting these values in the equation we get

Distance=\(\sqrt{(0-(-4))^{2}+(0 - 2)^{2}}\)

So we get

Distance of point P = √(16 + 4) = √20 = 2√5

Distance of point P = 2√5 units

Therefore, the distance of the point P (-4, 2) from the origin is 2√5 units.

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

NCERT Exemplar Class 10 Maths Exercise 7.1 Problem 3

The distance of the point P (–6, 8) from the origin is a. 8, b. 2√7, c. 10, d. 6

Summary:

The distance of the point P (–6, 8) from the origin is 10 units

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