Question

if the distance Between the points (x-,1) and (3,2) is 5 then the value of x is ​

Answer 1

Step-by-step explanation:

distance formula=√(x2-x1)²+(y2-y1)²

5=√(3-x)²+(3+1)²

25=(3-x)²+16

(3-x)²=25-16

(3-x)²=9

3-x=±3

3=x+3 or 3=x-3

x=3-3 or x=3+3

x=0 or x=6

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Answer 2

Answer:

x= -1 and 7

Step-by-step explanation:

Here let (×,-1) be x1,y1 ;(3,2) be ×2,y2 and 5 be x

(×)distance formula= under root( x2-x1)^2+(y2-y1)^2

5=under root (3-x1)^2+( 2-(-1))^2

squaring both side we get

25= (3-×1)^2 +3^2

25=9+x1^2-6x1+9

25-9-9 =x1^2-6×1

7=x1^2-6×1

x1^2-6x1-7=0

x1^2 -7×1 +×1 -7.....splitting the middle term

x1 (×1-7)+1 (×1-7)=0

(×1-7)(×1+1)=0

×1=7 and x1 = -1