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Mathematics

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.

Coordinate Geometry

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Answer

By formula,

Distance between two points (D) = (y2−y1)2+(x2−x1)2\sqrt{(y2 - y1)^2 + (x2 - x1)^2}

Given,

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

∴ PQ = QR

Substituting values we get :

⇒(−3−1)2+(5−0)2=(6−1)2+(x−0)2⇒(−4)2+52=52+x2⇒16+25=x2+25⇒x2+25=41\Rightarrow \sqrt{(-3 - 1)^2 + (5 - 0)^2} = \sqrt{(6 - 1)^2 + (x - 0)^2} \\[1em] \Rightarrow \sqrt{(-4)^2 + 5^2} = \sqrt{5^2 + x^2} \\[1em] \Rightarrow \sqrt{16 + 25} = \sqrt{x^2 + 25} \\[1em] \Rightarrow \sqrt{x^2 + 25} = \sqrt{41}

Squaring both sides we get :

⇒x2+25=41⇒x2=41−25⇒x2=16⇒x=16⇒x=±4.\Rightarrow x^2 + 25 = 41 \\[1em] \Rightarrow x^2 = 41 - 25 \\[1em] \Rightarrow x^2 = 16 \\[1em] \Rightarrow x = \sqrt{16} \\[1em] \Rightarrow x = \pm 4.

When x = 4,

P = (5, -3), Q = (0, 1) and R = (4, 6).

QR=(6−1)2+(4−0)2=52+42=25+16=41.PR=[6−(−3)]2+(4−5)2=[6+3]2+(−1)2=92+1=81+1=82.QR = \sqrt{(6 - 1)^2 + (4 - 0)^2} \\[1em] = \sqrt{5^2 + 4^2} \\[1em] = \sqrt{25 + 16} \\[1em] = \sqrt{41}. \\[1em] PR = \sqrt{[6 - (-3)]^2 + (4 - 5)^2} \\[1em] = \sqrt{[6 + 3]^2 + (-1)^2} \\[1em] = \sqrt{9^2 + 1} \\[1em] = \sqrt{81 + 1} \\[1em] = \sqrt{82}.

When x = -4,

P = (5, -3), Q = (0, 1) and R = (-4, 6).

QR=(6−1)2+(−4−0)2=52+(−4)2=25+16=41.PR=[6−(−3)]2+(−4−5)2=[6+3]2+(−9)2=92+(−9)2=81+81=162=92.QR = \sqrt{(6 - 1)^2 + (-4 - 0)^2} \\[1em] = \sqrt{5^2 + (-4)^2} \\[1em] = \sqrt{25 + 16} \\[1em] = \sqrt{41}. \\[1em] PR = \sqrt{[6 - (-3)]^2 + (-4 - 5)^2} \\[1em] = \sqrt{[6 + 3]^2 + (-9)^2} \\[1em] = \sqrt{9^2 + (-9)^2} \\[1em] = \sqrt{81 + 81} \\[1em] = \sqrt{162} \\[1em] = 9\sqrt{2}.

Hence, x = ±4,QR=41,PR=82,92.\pm 4, QR = \sqrt{41}, PR = \sqrt{82}, 9\sqrt{2}.

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