Question

if p(2, 5) q(x, -7) and pq=13 find the possible value of x​

Answer 1

Answer:

The possible values of x are 7 and –3

Step-by-step explanation:

The distance between two points A(x₁, y₁) and B(x₂, y₂) is given by,

$\boxed{ \tt AB=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}$

Given, the distance between the two points p(2,5) and q(x, -7) is 13 units.

Substitute :

x₁ = 2

y₁ = 5

x₂ = x

y₂ = -7

pq = √[(x – 2)² + (–7 – 5)²]

13 = √[(x – 2)² + (–12)²]

13² = (x – 2)² + (–12)²

169 = (x – 2)² + 144

(x – 2)² = 169 – 144

(x – 2)² = 25

x – 2 = √25

x – 2 = ±5

• x – 2 = 5

x = 5 + 2

x = 7

• x – 2 = –5

x = –5 + 2

x = –3

The possible values of x are 7 and –3