Question

 If the distance between the points (2,-3) and (10,p) is 10 , then value of p is *

3,9

-3,9

3,-9

-3,-9

​

Answer 1

Answer :-

➞ p = 3

Explanation :-

It is given to us that the distance between two points A and B is 10.

• A = (2, -3)
• B = (10, p)

And we have to find the value of p.

Comparing the given coordinates with (x, y), we get

• x₁ = 2 ; y₁ = -3
• x₂ = 10 ; y₂ = p

In other words, we have to find the ordinate of the second point B.

Using the distance formula, we have

⇒ AB² = (x₂ - x₁)² + (y₂ - y₁)²

For simplicity, we squared the distance formula and so the distance, so it doesn't matter whether you write it as :

• AB = √{ (x₂ - x₁)² + (y₂ - y₁)² }

OR

• AB² = (x₂ - x₁)² + (y₂ - y₁)²

Anyways, Let us move forward.

⇒ 10² = (10 - 2)² + (p - (-3))² [ AB = 10 ]

⇒ 100 = 8² + (p + 3)²

⇒ 100 = 64 + (p + 3)²

⇒ 36 = (p + 3)²

⇒ (6)² = (p + 3)²

If you carefully observe, The exponents of the base is the same. which means the base must also be same as they are equated by the equality sign here.

⇒ 6 = p + 3

⇒ p = 3

∴ The value of p is 3.

Answer 2

Answer:

• 1st Point = (2 , -3) = (x₁ , y₁)
• 2nd Point = (10 , p) = (x₂ , y₂ )
• Distance b/w points = 10

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf (Distance)^2=(x_2-x_1)^2+(y_2-y_1)^2\\\\\\:\implies\sf (10)^2=(10-2)^2+(p-(-3))^2\\\\\\:\implies\sf 100 = (8)^2+(p + 3)^2\\\\\\:\implies\sf 100=64 + (p + 3)^2\\\\\\:\implies\sf 100 - 64 = (p + 3)^2\\\\\\:\implies\sf 36 = (p + 3)^2\\\\\\:\implies\sf \sqrt{36} = p + 3\\\\\\:\implies\sf \pm \:6 = p + 3\\\\\\:\implies\sf p = 6 - 3 \quad or, \quad p = - \:6 - 3\\\\\\:\implies\underline{\boxed{\sf p = 3\quad or, \quad - \:9}}$

⠀

$\therefore\:\underline{\textsf{Hence, Required value of p is C) \textbf{3 or - 9}}}.$