If the distance between the points ( x,2) and (3,-6) is 10 units , then find the positive value of x

If the distance between the points ( x,2) and (3,-6) is 10 units , the

1.	If the distance between the points ( x,2) and (3,-6) is 10 units , then find the positive value of x
Answer» Distance between the points P(x, 2) and Q(3, - 6) is 10.Using distance formula PQ = 10{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt { ( x - 3 ) ^ { 2 } + \\{ 2 - ( - 6 ) \\} ^ { 2 } }{/tex}\xa0= 10{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt { x ^ { 2 } + 3 ^ { 2 } - 2 \\times 3 \\times x + ( 2 + 6 ) ^ { 2 } }{/tex}\xa0= 10{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt { x ^ { 2 } + 9 - 6 x + 8 ^ { 2 } }{/tex}\xa0= 10{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt { x ^ { 2 } + 9 - 6 x + 64 }{/tex}\xa0= 10{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt { x ^ { 2 } - 6 x + 73 }{/tex}\xa0= 10Squaring both sides, we getx2 - 6x + 73 = 100{tex}\\Rightarrow{/tex}\xa0x2 - 6x + 73 - 100 = 0{tex}\\Rightarrow{/tex}\xa0x2- 6x -27 = 0{tex}\\Rightarrow{/tex}\xa0x2- 9x + 3x-2 7 = 0{tex}\\Rightarrow{/tex}\xa0x(x - 9) + 3(x - 9) = 0{tex}\\Rightarrow{/tex} (x - 9) (x + 3) = 0{tex}\\Rightarrow{/tex}\xa0either\xa0x-9 = 0 or x + 3 = 0{tex}\\Rightarrow{/tex}\xa0x = 9 or x = - 3Ignoring x = - 3 as it is given that x is a positive integer.Thus, only solution\xa0is x = 9.