Give me the best solution of coordinate geometry of ncert 10 class . How i solve this question fast and easy way please help me to solve this important question. 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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# Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4). Q.10

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Point (x, y) is equidistant from (3, 6) and ( – 3, 4).

Squaring both sides, (x – 3)

^{2}+(y – 6)^{2}= (x + 3)^{2}+(y – 4)^{2}x

^{2 }+ 9 – 6x + y^{2}+ 36 – 12y = x^{2 }+ 9 + 6x + y^{2 }+16 – 8y36 – 16 = 6x + 6x + 12y – 8y

20 = 12x + 4y

3x + y = 5

3x + y – 5 = 0