Question

If Q ( 0,1) is a equidistant from P(5,-3) and R (x,6) and find the values of x . Also find the distance QR and PR .

Hint : find by the distance formula .

Answer 1

According to question,

QP = QR

=> sqrt [ ( 0 - 5)^2 + ( 1 +3)^2] = sqrt [ (0 - x)^2 + (1 - 6)^2]

On squaring both sides, we get

=> ( - 5)^2 + (4)^2 = ( - x)^2 + (-5)^2

=> 25 + 16 = x^2 + 25

=> x^2 = 16

=> x = 4



Now,

QR = sqrt [ ( 0 + 4)^2 + ( 1 - 6)^2]

= sqrt ( 16 + 25)

= sqrt ( 41)



PR = sqrt [ (5 + 4)^2 + ( - 3 - 6)^2]

= sqrt ( 81 + 81)

= sqrt ( 162)

= 9 sqrt ( 2)

Answer 2

As we know that :-

Distance formula is :-

$= > d = \sqrt{( {x _{2} }^{2} - {x _{1} }^{2} ) + ( {y _{2} }^{2} - {y _{1} }^{2} )}$

_______________________________

It is the given that point Q(0,1) is equal distance from point P(5,-3) and point (X, 6).

then

By distance formula.. and according to condition we get :-

=> PQ = QR

$= > \sqrt{ ({0}^{2} - {5}^{2}) + ( {1}^{2} - ( { - 3}^{2}) } \\ = \sqrt{ ({x}^{2} - {0}^{2} ) +( {6}^{2} - {1}^{2} )}$