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
