If the distance between the points (4, p) and (1, 0) is 5

Q) If the distance between the points (4, p) and (1, 0) is 5, what is the value of p?

(Q 21 – 30/2/3 – CBSE 2026 Question Paper)

Ans:

Step 1: We know that the distance between the two points (x₁,y₁) and (x₂, y₂) is given by:
$\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Step 2: Here, we have points (4, p) and (1, 0)

∴ $\sqrt{(1-4)^2 + (0-p)^2}$ = 5

∴ $\sqrt{(-3)^2 + (-p)^2}$ = 5

∴ $\sqrt{9 + p^2}$ = 5

∴ $(\sqrt{9 + p^2})^2 = (5)^2$

∴ 9 + p ² = 25

∴ p ² = 25 -9 = 16

∴ p = √16 = 4

Therefore, the value of p is 5.

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