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Coordinate Geometry
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08/06/2021 2:42 pm
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Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
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08/06/2021 2:44 pm
Point (x, y) is equidistant from (3, 6) and (-3, 4).
= \(\sqrt{(x-3)^2 + (y-6)^2}\)
= \(\sqrt{(x-(-3))^2 + (y-4)^2}\)
= \(\sqrt{(x-3)^2 + (y-6)^2}\)
= \(\sqrt{(x+3))^2 + (y-4)^2}\)
Squaring both sides, (x – 3)2+(y – 6)2 = (x + 3)2 +(y – 4)2
x2 + 9 – 6x + y2+ 36 – 12y = x2 + 9 + 6x + y2 +16 – 8y
36 – 16 = 6x + 6x + 12y – 8y
20 = 12x + 4y
3x + y = 5
3x + y – 5 = 0