Question

find value of a if distance between A (-3,-14) and B (a,-5) is 9 units.

Answer 1

We have :

A ( -3 , -14 ) and B ( a , - 5 ).

Here,

X1 = -3 , Y1 = -14 and X2 = a , Y2 = -5

Given that : Distance between point A ( -3 , -14 ) to Point B ( a , - 5 ) is 9 units.

AB = 9 => AB² = 81

=> ( X2 - X1 )² + ( Y2 - Y1 )² = 81

=> ( a + 3 )² + ( -5 + 14 )² = 81

=> (a)² + (3)² + 2a × 3 + 81 = 81

=> a² + 9 + 6a = 0

=> a² + 6a + 9 = 0

=> a² + 3a + 3a + 9 = 0

=> a ( a + 3 ) + 3 ( a + 3 ) = 0

=> ( a + 3 ) ( a + 3 ) = 0

=> ( a + 3 ) = 0

=> a = -3.

Answer 2

⇒$\sqrt{(-3-a)^2+(-14-(-5))^2} =9$

⇒$\sqrt{((3^2+a^2+6a)+(-9)^2} =9$

⇒$\sqrt{((3^2+a^2+6a)+81} =9$

Squaring Both the Sides

⇒$((3^2+a^2+6a)+81 =81$

⇒$a^2+6a+9=0$

⇒$a^2 + 3a+3a + 9=0$

⇒$(a+3a)^2=0$

a=-3