Question

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

Answer 1

QP = QR

QP^2 = QR^2

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

=> 25 + 16 = x^2 + 25

=> x^2 = 16

=> x = ± 4

Hence R is (4 , 6) or (- 4 , 6)

Now  

QR (0 , 1)(4 , 6)

QR = $\bf\huge\sqrt{(4 - 0)^2 + (6 - 1)^2}$

QR = $\bf\huge\sqrt{(16 + 25)}$

QR = $\bf\huge\sqrt{41}$

Also

QR (0 , 1) (-4 , 6)

QR = $\bf\huge\sqrt{(- 4 - 0)^2 + (6 - 1)^2}$

QR = $\bf\huge\sqrt{16 + 25}$

QR = $\bf\huge\sqrt{41}$

Now

PR (5 , -3)(4 , 6)

PR = $\bf\huge\sqrt{(4 - 5)^2 + (6 + 3)^2}$

PR = $\bf\huge\sqrt{1 + 81}$

PR = $\bf\huge\sqrt{82}$

Also

PR (5 , -3)(-4 , 6)

PR = $\bf\huge\sqrt{(-4 - 5)^2 + (6 + 3)^2}$

PR = $\bf\huge\sqrt{81 + 81}$

PR = $\bf\huge 9 \sqrt{2}$