Question

P(-6,-3), Q(-1,9) Find the distances between the points.

Answer 1

Let the Points P(-6,-3) and Q(-1,9) be P(x₁,y₁) and Q(x₂, y₂) respectively. 

Using the Distance Formula, 
   $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
⇒ $d = \sqrt{(-1 + 6)^2 + (9 + 3)^2}$
⇒ $d = \sqrt{(5)^2 + (12)^2}$
⇒ $d = \sqrt{(25 + 144)}$
⇒ $d = \sqrt{169}$
∴ d = 13


Hence, the Distance between the two points is 13 units. 


Hope it helps.

Answer 2

SOLUTION:-
GIVEN BY POINTS :-
P( -6, -3) , Q( -1 , 9)
HERE ,
$x _{1} = - 6 \\ x _{2} = - 1 \\ y _{1} = - 3 \\ y _{2} = 9$
using formul for distance between two points.
$d = \sqrt{( {x _{2} - x _{1}) }^{2} + {( {y _{2} - y _{1}) }^{2} } } \\ d = \sqrt{ ({ - 1 + 6)}^{2} + { (9 + 3)}^{2} } \\ d = \sqrt{ {5}^{2} + {12}^{2} } \\ d = \sqrt{25 + 144} \\ d = \sqrt{169} \\ d = 13$
hence, distances between the points. =13 unit

■ I HOPE ITS HELP■