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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9

The point (x, y) is equidistant from the points (3, 4) and (−5, 6). Find a relation between x and y - Mathematics

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Question

The point (x, y) is equidistant from the points (3, 4) and (−5, 6). Find a relation between x and y

Sum

Solution

Let the point O be (x, y), A be (3, 4) and B be (−5, 6).

Distance = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

Given, OA = OB

`sqrt((x - 3)^2 + (y - 4)^2) = sqrt((x + 5)^2 + (y - 6)^2`

Squaring on both sides

(x – 3)2 + (y – 4)2 = (x + 5)2 + (y – 6)2

x2 – 6x + 9 + y2 – 8y + 16 = x2 + 10x + 25 + y2 – 12y + 36

x2 + y2 – 6x – 8y + 25 = x2 + y2 + 10x – 12y + 61

6x – 10x – 8y + 12y = 61 – 25

⇒ – 16x + 4y = 36 ........(÷ 4)

⇒ – 4x + y = 9

∴ The relation between x and y is y = 4x + 9

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Co-ordinates of Points and Distance
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Chapter 5: Coordinate Geometry - Exercise 5.2 [Page 203]

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Samacheer Kalvi Mathematics [English] Class 9 TN Board
Chapter 5 Coordinate Geometry
Exercise 5.2 | Q 9 | Page 203
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