Question

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

Answer 1

PQ = QR

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

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

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

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

= `sqrt(25+16) `

= `sqrt(x^2+25)`

41 = x² + 25

16 = x²

x = ±4

Therefore, point R is (4, 6) or (−4, 6).

When point R is (4, 6),

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

= `sqrt((1^2+(-9)^2)) `

= `sqrt(1+81)`

= `sqrt82`

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

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

= `sqrt(16+25)`

= `sqrt41`

When point R is (−4, 6),

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

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

= `sqrt(81+81)`

= `9sqrt2`

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

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

= `sqrt(16+25)`

= `sqrt41`

Answer 2

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 = x² + (-5)²

41 = x² + 25

x² = 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