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Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.

Coordinate Geometry

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Answer

By distance formula,

d = (x2−x1)2+(y2−y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

Given, PQ = QR.

∴ Distance between (6, -1) and (1, 3) = Distance between (1, 3) and (x, 8).

⇒(1−6)2+[3−(−1)]2=(x−1)2+(8−3)2⇒(−5)2+[4]2=x2+1−2x+(5)2⇒25+16=x2−2x+1+25⇒41=x2−2x+26\Rightarrow \sqrt{(1 - 6)^2 + [3 - (-1)]^2} = \sqrt{(x - 1)^2 + (8 - 3)^2} \\[1em] \Rightarrow \sqrt{(-5)^2 + [4]^2} = \sqrt{x^2 + 1 - 2x + (5)^2} \\[1em] \Rightarrow \sqrt{25 + 16} = \sqrt{x^2 - 2x + 1 + 25} \\[1em] \Rightarrow \sqrt{41} = \sqrt{x^2 - 2x + 26}

On squaring both sides,

⇒41=x2−2x+26⇒x2−2x+26−41=0⇒x2−2x−15=0⇒x2−5x+3x−15=0⇒x(x−5)+3(x−5)=0⇒(x+3)(x−5)=0⇒x=−3 or x=5.\Rightarrow 41 = x^2 - 2x + 26 \\[1em] \Rightarrow x^2 - 2x + 26 - 41 = 0 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x^2 - 5x + 3x - 15 = 0 \\[1em] \Rightarrow x(x - 5) + 3(x - 5) = 0 \\[1em] \Rightarrow (x + 3)(x - 5) = 0 \\[1em] \Rightarrow x = -3 \text{ or } x = 5.

Hence, x = -3 or 5.

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