Question

The distance between the points P(1,3) and Q(x,7) is 5 units. Find the possible values of x.​

Answer 1

x = - 2 , x = 4

Explanation:

Calculate the distance PQ using the distance formula and equate to 5

d =
`\sqrt{(x_(2)-x_(1))^2+(y_(2)-y_(1))^2 }`

with (x₁, y₁ ) = P (1, 3) and (x₂, y₂ ) = Q (x, 7 )

PQ =
`√((x-1)^2+(7-3)^2)` =
`√((x-1)^2+4^2)` =
`√((x-1)^2+16)` , then

`√((x-1)^2+16)` = 5 ( square both sides )

(x - 1)² + 16 = 25 ( subtract 16 from both sides )

(x - 1)² = 9 ( take square root of both sides )

x - 1 = ±
`√(9)` = ± 3 ( add 1 to both sides )

x = 1 ± 3

Then

x = 1 - 3 = - 2 , or

x = 1 + 3 = 4