Q.

If the distance between the points (3, –5) and (x, –5) is 15 units, then the values of x are :

1 12, –18  
2 –12, 18  
3 18, 5  
4 –9, –12  

Ans.

(2)

Here, (15)2=(3-x)2+(-5+5)2

⇒225=9-6x+x2⇒x2-6x-216=0

⇒x2-18x+12x-216=0⇒x(x-18)+12(x-18)=0

⇒(x-18)(x+12)=0⇒x=-12,18