Question

R(-3a,a), S(a,-2a) Find the distances between the points.

Answer 1

Let the Points R(-3a,a) and S(a,-2a) be R(x₁,y₁) and S(x₂, y₂) respectively. 

Using the Distance Formula, 
   $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
⇒ $d = \sqrt{(a + 3a)^2 + (-2a - a)^2}$
⇒ $d = \sqrt{(4a)^2 + (-3a)^2}$
⇒ $d = \sqrt{(16a^2 + 9a^2)}$
⇒ $d = \sqrt{25a^2}$
∴ d = 5a


Hence, the Distance between the two points is 5a units. 


Hope it helps.

Answer 2

SOLUTION:-
GIVEN POINTS :- R(-3a,a), S(a,-2a)
here,
$x _{1} = - 3a \: \: \: \:y _{1} = a\\ x _{2} = a \: \: \: \: y _{2} = - 2a\\ distance \: formula \: \\ d = \sqrt{ {(x _{2} - x _{1})}^{2} + {(y _{2} - y _{2} ) }^{2} } \\ d = \sqrt{ {(a + 3a)}^{2} + {( - 2a - a)}^{2} } \\ d = \sqrt{16 {a}^{2} + 9 {a}^{2} } \\ d = \sqrt{25 {a}^{2} } \\ d = 5$
hence, the distance is 5 unit


■I HOPE ITS HELP■