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

There isnt a lot of information provided....

The distance between the points P(1,3) and Q(x,7)

netflixacc0107

23.11.2021

x = - 2 , x = 4

Step-by-step 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 = $\sqrt{(x-1)^2+(7-3)^2}$ = $\sqrt{(x-1)^2+4^2}$ = $\sqrt{(x-1)^2+16}$ , then

$\sqrt{(x-1)^2+16}$ = 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 = ± $\sqrt{9}$ = ± 3 ( add 1 to both sides )

x = 1 ± 3

Then

x = 1 - 3 = - 2 , or

x = 1 + 3 = 4

0,0(0 оценок)