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Mathematics Question on Coordinate Geometry

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

PR = QR
(5−0)2+(−3−1)2=(0−x)2+(1−6)2\sqrt{(5-0)^2+(-3-1)^2}=\sqrt{(0-x)^2+(1-6)^2}
(5)2+(−4)2=(−x)2+(−5)2\sqrt{(5)^2+(-4)^2}=\sqrt{(-x)^2+(-5)^2}
25+16=x2+25\sqrt{25+16}=\sqrt{x^2+25}
41=x2+25x^2+25
x2=16x^2=16
x=±4x=\pm 4
Therefore, point R is (4,6) or (-4,6)

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

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