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Mathematics

Find a relation between x and y such that point (x, y) is equidistant from the points (7, 1) and (3, 5).

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,

∴ Distance between (x, y) and (7, 1) = Distance between (x, y) and (3, 5).

∴(7−x)2+(1−y)2=(3−x)2+(5−y)2⇒49+x2−14x+1+y2−2y=9+x2−6x+25+y2−10y\therefore \sqrt{(7 - x)^2 + (1 - y)^2} = \sqrt{(3 - x)^2 + (5 - y)^2} \\[1em] \Rightarrow \sqrt{49 + x^2 - 14x + 1 + y^2 - 2y} = \sqrt{9 + x^2 - 6x + 25 + y^2 - 10y}

On squaring both sides,

⇒49+x2−14x+1+y2−2y=9+x2−6x+25+y2−10y⇒x2+y2−14x−2y+50=x2+y2−6x−10y+34⇒x2+y2−x2−y2−6x+14x=10y−2y+50−34⇒8x=8y+16⇒8x=8(y+2)⇒x=y+2⇒x−y=2.\Rightarrow 49 + x^2 - 14x + 1 + y^2 - 2y = 9 + x^2 - 6x + 25 + y^2 - 10y \\[1em] \Rightarrow x^2 + y^2 - 14x - 2y + 50 = x^2 + y^2 - 6x - 10y + 34 \\[1em] \Rightarrow x^2 + y^2 - x^2 - y^2 - 6x + 14x = 10y - 2y + 50 - 34 \\[1em] \Rightarrow 8x = 8y + 16 \\[1em] \Rightarrow 8x = 8(y + 2) \\[1em] \Rightarrow x = y + 2 \\[1em] \Rightarrow x - y = 2.

Hence, relation between x and y is x - y = 2.

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