Question

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Answer 1

If Q (0,1) is equidistant from P(5,-3) and R( x,6) then distance between Q (0,1) and P(5,-3) is equal to distance between Q (0,1) and R(x,6).

Let distance between Q (0,1) and P(5,-3) be d.

Now , we will use distance formula to find out distance d :-

➝ d²= (5-0)²+ (-3-1)²

➝ d²= 25+ (-1)² (3+1)²

➝ d²= 25+ 16

➝ d²= 41

Let distance between Q (0,1) and R(x,6) be t.

Now , we will use distance formula to find out distance t :-

➝ t² = (x-0)²+ (6-1)²

➝ t² = x²+ (5)²

➝ t² = x²+ 25

➡️Now, distance between Q (0,1) and P(5,-3) = distance between Q (0,1) and R(x,6)

➡️ d = t

➡️ d² = t²

➝ 41 = x² + 25

➝ 41-25 = x²

➝ 16 = x²

➝ ±4 = x

Answer :

value of x = ±4

Answer 2

$\large\purple{ \mathcal{\underline{Given :-}}}$

• Q (0 , 1) is equidistant from P (5 , -3) and R (x , 6).

$\large\pink{ \mathcal{\underline{To \: Find :-}}}$

• The value of x

$\large\orange{ \mathcal{\underline{Solution :-}}}$

The distance between Q and P and the distance between Q and R are same.

So, Let the distance between Q and P be d

We know that,

$\bigstar \: \bf \: Distance = \sqrt{ (x_{2} - x_{1} )^{2} + {( y_{2} - y_{1} )}^{2} }$

$\therefore \: d = \sqrt{ {(5 - 0)}^{2} + {( - 3 - 1)}^{2} }$

$\: \: \: \: \: \: \: \: \: = \sqrt{ {5}^{2} + ( - 4) ^{2} }$

$\: \: \: \: \: \: \: \: \: = \sqrt{25 + 14}$

$\: \: \: \: \: \: \: \: \: = \sqrt{41}$

Now let the distance between Q and R be d'

$\therefore \: d' = \sqrt{(x - 0)^{2} + {(6 - 1)}^{2} }$

$\: \: \: \: \: \: \: \: \: \: = \sqrt{ {x}^{2} + {5}^{2} }$

$\: \: \: \: \: \: \: \: \: \: = \sqrt{ {x}^{2} + 25 }$

$Hence,$

$d' = d$

$\rightarrow \sqrt{ {x}^{2} + 25 } = \sqrt{41}$

$\rightarrow \:{x}^{2} + 25 = 41$

$\rightarrow \: {x}^{2} = 41 - 25$

$\rightarrow \: {x}^{2} = 16$

$\rightarrow \: \sqrt{ {x}^{2} } = \pm \sqrt{16}$

$\rightarrow \: x = \pm \: 4$

Distance can not be negative.

$\sf\green{Therefore, \: the \: Value \: of \: x \: is \: 4. }$