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Mathematics

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).

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,

Point (x, y) is equidistant from the point (3, 6) and (-3, 4).

⇒(y−6)2+(x−3)2=(y−4)2+[x−(−3)2]⇒y2+36−12y+x2+9−6x=y2+16−8y+[x+3]2⇒y2+x2−6x−12y+45=y2+16−8y+x2+9+6x⇒x2+y2−12y−6x+45=x2+y2+6x−8y+25\Rightarrow \sqrt{(y - 6)^2 + (x - 3)^2} = \sqrt{(y - 4)^2 + [x - (-3)^2]} \\[1em] \Rightarrow \sqrt{y^2 + 36 - 12y + x^2 + 9 - 6x} = \sqrt{y^2 + 16 - 8y + [x + 3]^2} \\[1em] \Rightarrow \sqrt{y^2 + x^2 - 6x - 12y + 45} = \sqrt{y^2 + 16 - 8y + x^2 + 9 + 6x} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 12y - 6x + 45} = \sqrt{x^2 + y^2 + 6x - 8y + 25}

Squaring both sides we get :

⇒x2+y2−12y−6x+45=x2+y2+6x−8y+25⇒x2−x2+y2−y2+6x+6x−8y+12y=45−25⇒12x+4y=20⇒4(3x+y)=20⇒3x+y=5⇒3x+y−5=0.\Rightarrow x^2 + y^2 - 12y - 6x + 45 = x^2 + y^2 + 6x - 8y + 25 \\[1em] \Rightarrow x^2 - x^2 + y^2 - y^2 + 6x + 6x - 8y + 12y = 45 - 25 \\[1em] \Rightarrow 12x + 4y = 20 \\[1em] \Rightarrow 4(3x + y) = 20 \\[1em] \Rightarrow 3x + y = 5 \\[1em] \Rightarrow 3x + y - 5 = 0.

Hence, 3x + y - 5 = 0.

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