Question

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

Answer 1

Answer:-

Given:

Distance between (4 , p) & (1 , 0) = 5 units.

We know that,

Distance between two points with coordinates (x₁ , y₁) & (x₂ , y₂) is :

$\sf \large{\sqrt{(x_2 - x_1)^2 - (y_2 - y_1)^2}}$

Let,

• x₁ = 4
• y₁ = p
• x₂ = 1
• y₂ = 0

Hence,

$\implies \sf \sqrt{(1 - 4) ^{2} + (0 - p) ^{2} } = 5 \\ \\ \implies \sf \: \sqrt{( - 3) ^{2} + {( - p)}^{2} } = 5$

On squaring both sides we get,

$\: \implies \sf \: {( - 3)}^{2} + {( - p)}^{2} = 25 \\ \\ \implies \sf \: 9 + {p}^{2} = 25 \\ \\ \implies \sf \: {p}^{2} = 25 - 9 \\ \\ \implies \sf \: {p}^{2} = 16 \\ \\ \implies \sf \: p \: = \sqrt{16} \\ \\ \implies \sf \red{p = \pm \: 4}$

∴ The value of p is ± 4.

Answer 2

Answer :-

• Value of p is ±4 units.

Given :-

• Distance between the points (4, p) and (1, 0)

To Find :-

• value of p.

Solution :-

Here,

• x1 = 4
• y1 = p
• x2 = 1
• y2 = 0

→ (x1, y1) = (4, p)

→ (x2, y2) = (1, 0)

Formula for finding distance between two points is

√ { (x2 - x1)² + (y2 - y1)² }

Put the values in the formula

→ √ { (1 - 4)² + (0 - p)² } = 5

→ √ { (-3)² + (-p)² } = 5

→ 9 + p² = 25

→ p² = 25 - 9

→ p² = 16

→ p = √16

→ p = ±4 units

Hence, the value of p is ±4 units.