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.
x = ±4
QR = \(\sqrt{41}\)
PR = \(\sqrt{82}, 9\sqrt{2}\)
We are given that Q(0, 1) is equidistant from P(5, −3) and R(x, 6). This means the distances QP and QR are equal.
First, find QP using the distance formula between \((x_1, y_1)\) and \((x_2, y_2)\): \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
For Q(0, 1) and P(5, −3): \( QP = \sqrt{(0 - 5)^2 + (1 - (−3))^2} = \sqrt{(−5)^2 + (4)^2} = \sqrt{25 + 16} = \sqrt{41} \).
Now find QR in terms of x for Q(0, 1) and R(x, 6): \( QR = \sqrt{(0 - x)^2 + (1 - 6)^2} = \sqrt{x^2 + (−5)^2} = \sqrt{x^2 + 25} \).
Since QP and QR are equal: \( \sqrt{41} = \sqrt{x^2 + 25} \).
Squaring both sides: \( 41 = x^2 + 25 \).
So, \( x^2 = 41 − 25 = 16 \) which gives \( x = \pm 4 \).
Next, find QR using \( x = 4 \) or \( x = −4 \). In both cases, \( QR = \sqrt{x^2 + 25} = \sqrt{16 + 25} = \sqrt{41} \).
Now find PR when \( x = 4 \). Then R is (4, 6). Distance between P(5, −3) and R(4, 6): \( PR = \sqrt{(4 − 5)^2 + (6 − (−3))^2} = \sqrt{(−1)^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82} \).
When \( x = −4 \), R is (−4, 6). Distance between P(5, −3) and R(−4, 6): \( PR = \sqrt{(−4 − 5)^2 + (6 − (−3))^2} = \sqrt{(−9)^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2} \).
Thus, \( x = \pm 4 \), \( QR = \sqrt{41} \), and \( PR = \sqrt{82} \) or \( 9\sqrt{2} \) for the two positions of R.