Question

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.

Answer 1

PR = 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=\(x^2+25\)
\(x^2=16\)
\(x=\pm 4\)
Therefore, point R is (4,6) or (-4,6)

When point R is (4,6)
PR= \(\sqrt{(5-4)^2+(-3-6)^2}=\sqrt{(1)^2+(-9)^2}=\sqrt{1+81}=\sqrt{82}\)
QR= \(\sqrt{(0-4))^2+(1-6)^2}=\sqrt{(4)^2+(-5)^2}=\sqrt{16+25}=\sqrt{41}\)

When point R is (-4,6)
PR= \(\sqrt{(5-(-4))^2+(-3-6)^2}=\sqrt{(9)^2+(-9)^2}=\sqrt{81+81}=\sqrt{162}\)
QR= \(\sqrt{(0-(-4))^2+(1-6)^2}=\sqrt{(4)^2+(-5)^2}=\sqrt{16+25}=\sqrt{41}\)