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

Given:

Points Q(0,1) is equidistant from P(5,-3) and R(x,6)

So,

∴ QP = QR

⇒ QR² = QR²  [By squaring both the sides]

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

[As the distance formula = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ ]

⇒ 5² + 4² = x² + 5²

⇒ 25 + 16  = x² + 25

⇒ x² = 16

⇒ x = ±4

Thus,

The coordinates of R are (4,6) and (-4,6)

• CASE 1

QR = Distance between Q(0,1) and R(4,6)

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

      = $\sqrt{4^{2}+5^{2}}$

      = $\sqrt{41}\ units$

And

PR = Distance between P(5,-3) and R(4,6)

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

    = $\sqrt{ (-1)^{2} + (9)^{2} }$

    = $\sqrt{82}\ units$

• CASE 2

When the coordinates or R = (-4,6)

In this case,

QR = Distance between Q(0,1) and R(-4,6)

     = $\sqrt{(-4-0)^{2}+(6-1)^{2}}$

     = $\sqrt{(-4)^{2}+(5)^{2}}$

     = $\sqrt{41}\ units$

PR = Distance between P(5,-3) and R(-4,6)

     = $\sqrt{(-4-5)^{2}+(6+3)^{2}}$

     = $\sqrt{(-9)^{2}+(9)^{2}}$

     = $\sqrt{162} = 9\sqrt{2}\ units$