If the distance between the points (4, p) and (1, 0) is 5, then the value of p is
a. 4 only
b. ±4
c. –4 only
d. 0

Solution:

We know that

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

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

The points given are (4, p) and (1, 0)

Let us apply the distance formula

\(\\5=\sqrt{(4-1)^{2}+p^{2}}\)

By squaring on both sides

25 = (4 - 1)² + p²

25 = 3² + p²

By further calculation

25 = 9 + p²

p² = 25 - 9 = 16

So we get

p = ±4

Therefore, the value of p is ±4.

✦ Try This: If the distance between the points (5, q) and (2, 0) is 6, then the value of q is

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

NCERT Exemplar Class 10 Maths Exercise 7.1 Problem 19

If the distance between the points (4, p) and (1, 0) is 5, then the value of p is a. 4 only, b. ± 4, c. – 4 only, d. 0

Summary:

If the distance between the points (4, p) and (1, 0) is 5, then the value of p is ±4

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