Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).
Solution and Explanation
Point (x, y) is equidistant from (3, 6) and (−3, 4).
\(\therefore\) \(\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x-(-3))^2+(y-4)^2}\)
\(\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}\)
\({(x-3)^2+(y-6)^2}={(x+3)^2+(y-4)^2}\)
\(36-16=6x+6x+12y-8y\)
\(20=12x+4y\)
\(3x+y=5\)
\(3x+y-5=0\)
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Concepts Used:
Coordinate Geometry
The central idea in coordinate geometry is to assign numerical coordinates to points in a plane or space, which allows us to describe their positions and relationships using algebraic equations. The most common coordinate system is the Cartesian coordinate system, named after the French mathematician and philosopher René Descartes.