Question

if distance between the point (x,1) and (1,x) is √2 then value of x is​

Answer 1

Answer:

hi

Step-by-step explanation:

value of x is 7 and -1

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Answer 2

Answer:

value of 'x' is '0' or '2'

Step-by-step explanation:

we have to find the value of 'x' if the  distance between the points (x, 1) and (1, x)  is √2

distance formula is $\sqrt{(a-p)^{2} +(b-y)^{2} }$

where (a,b) and (p,y) are given coordinates

let's insert the values:

a = x, b = 1, p = 1, y = x

distance is √2

equate it;

$\sqrt{(x-1)^{2}+(1-x)^{2} }$= √2

squaring on both sides of above equation , we get:

(x - 1)² + (1 - x)² = (√2)²

(x - 1)² + (1 - x)²  =  (√2) (√2)= 2

we know that (a - b)² = a² + b² - 2ab

x² + 1² - 2(1)(x) + 1² + x² - (2)(1)(x) = 2

x² + 1 - 2x + 1 + x² - 2x = 2

2x² + 2 - 4x = 2

2x² - 4x = 0

take '2x' common from above

(2x)(x - 2) = 0

2x = 0 or (x - 2) =0

x = 0 or x = 2

value of 'x' is '0' or '2'