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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

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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
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