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Mathematics · Coordinate Geometry · NCERT Examples

Example 5

A complete, independently verified solution for NCERT Example 5.

Mathematics · Chapter 7 · NCERT Examples

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NCERT Example

Example 5

y-axis · equidistant point · distance formula

Find a point on the y-axis which is equidistant from A(6, 5) and B(-4, 3).

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:

A point on the y-axis has x-coordinate 0. Let the required point be P(0, y).

Since P is equidistant from A and B:

AP=BPAP=BP

Squaring both sides:

(6−0)2+(5−y)2=(−4−0)2+(3−y)2(6-0)^2+(5-y)^2=(-4-0)^2+(3-y)^2

Expanding:

36+25+y2−10y=16+9+y2−6y36+25+y^2-10y=16+9+y^2-6y

Cancelling y² and simplifying:

61−10y=25−6y61-10y=25-6y
36=4y36=4y
y=9y=9

Check:

AP=(6−0)2+(5−9)2=36+16=52AP=\sqrt{(6-0)^2+(5-9)^2}=\sqrt{36+16}=\sqrt{52}
BP=(−4−0)2+(3−9)2=16+36=52BP=\sqrt{(-4-0)^2+(3-9)^2}=\sqrt{16+36}=\sqrt{52}

Thus AP = BP.

Final answer

The required point is (0, 9).