Question:
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).
Updated On: Jan 13, 2026
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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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