# If Q (0, 1) is Equidistant from P (5, -3) and R (X, 6), Find the Values of X. Also, Find the Distances Qr and Pr - Mathematics

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

#### Solution

The distance d between two points (x_1,y_1) and (x_2,y_2) is given by the formula

d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)

The three given points are Q(0,1), P(5,3) and R(x,6).

Now let us find the distance between 'P' and 'Q'.

PQ = sqrt((5 - 0)^2 + (-3-1)^2)

= sqrt((5)^2 + (-4)^2)

= sqrt(25 + 16)

PQ = sqrt(41)

Now, let us find the distance between ‘Q’ and ‘R’.

QR = sqrt((0 - x)^2 + (1- 6)^2)

QR = sqrt((-x)^2 + (-5)^2)

It is given that both these distances are equal. So, let us equate both the above equations,

PQ= QR

sqrt(41) = sqrt((-x)^2 + (-5)^2)

Squaring on both sides of the equation we get,

41 = (-x)^2 + (-5)^2

41 = x^2 + (-5)^2

41 = x^2 + 25

x^2 = 16

x = +-4

Hence the values of ‘x’ are 4 or (-4)

Now, the required individual distances,

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

sqrt((+-4)^2 + (-5)^2)

= sqrt(16 + 25)

QR = sqrt(41)

Hence the length of ‘QR’ is sqrt(41) units

For ‘PR’ there are two cases. First when the value of ‘x’ is 4,

PR = sqrt(82)

Then when the value of ‘x’ is -4,

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

=sqrt((9)^2 + (-9)^2)

= sqrt(81 + 81)

PR = 9sqrt2

Hence the length of 'PR' can be sqrt(82) or 9sqrt(2)units

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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 34 | Page 16