Mathematics · Coordinate Geometry · NCERT Examples

Example 4

A complete, independently verified solution for NCERT Example 4.

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Mathematics · Chapter 7 · NCERT Examples

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NCERT Example

Example 4

equidistant point · distance formula · perpendicular bisector

Find a relation between x and y such that P(x, y) is equidistant from A(7, 1) and B(3, 5).

Coordinate axes showing A(7, 1), B(3, 5), and their perpendicular bisector x minus y equals 2.
Figure 7-7 from the supplied NCERT chapter.

Solution

Board-exam working:

Since P is equidistant from A and B:

AP=BPAP=BP

Squaring both sides:

AP2=BP2AP^2=BP^2
(x−7)2+(y−1)2=(x−3)2+(y−5)2(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2

Expanding both sides:

x2−14x+49+y2−2y+1=x2−6x+9+y2−10y+25x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25

Cancelling equal terms and simplifying:

−14x−2y+50=−6x−10y+34-14x-2y+50=-6x-10y+34
8y−8x+16=08y-8x+16=0
x−y=2x-y=2

Final answer

The required relation is x - y = 2.

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