Question

the distance the points (4, p) and (1,o) is 5 units then the value of p ​

Answer 1

$\huge\underline\mathbb{\red Q\pink{U}\purple{ES} \blue{T} \orange{IO}\green{N :}}$

The distance between the points (4, p) and (1,0) is 5 units then the value of p is..?

$\huge\underline\mathbb{\red S\pink{O}\purple{LU} \blue{T} \orange{IO}\green{N :}}$

★ Given that,

• A(4, p)
• B(1, 0)
• Distance between AB = 5 units.

★ To find,

• Value of " p ".

★ Let,

• x1 = 4 ; y1 = p
• x2 = 1 ; y2 = 0

By using distance formula we can get the value of " p ".

$\tt\blue{AB = \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} }}$

• Substitute the values.

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

$\sf\:⟹ 5 = \sqrt{(-3)^{2} + (-p)^{2}}$

$\sf\:⟹ 5 = \sqrt{ 9 + p^{2}}$

• Squaring on both sides.

$\sf\:⟹{(5)}^{2} = {( \sqrt{9 + {p}^{2} }) }^{2}$

$\sf\:⟹ 25 = 9 + p^{2}$

$\sf\:⟹ p^{2} = 25 - 9$

$\sf\:⟹ p^{2} = 16$

$\sf\:⟹ p = \sqrt{16}$

$\sf\:⟹ p = 4$

$\underline{\boxed{\bf{\red{∴ The \: value \: of \: p = ± 4.}}}}\:\orange{\bigstar}</p><p>$

Answer 2

.......I hope It's helpful for you.........