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If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also find the length of AB.

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If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also, find the length of AB.

If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p.

Answer in Brief
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Solution 1

The given points are A(0, 2), B(3, p) and C(p, 5).

It is given that A is equidistant from B and C.

∴ AB = AC

⇒ AB2 = AC2

⇒ (3 − 0)2 + (p − 2)2 = (p − 0)2 + (5 − 2)2

⇒ 9 + p2 + 4 − 4p = p2 + 9

⇒ 4 − 4p = 0

⇒ 4p = 4

p = 1

Thus, the value of p is 1.

Length of AB `=sqrt((3-0)^2+(1-2)^2)=sqrt(3^2+(-1)^2)=sqrt(9+1)=sqrt(10) units`

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Solution 2

It is given that A(0, 2) is equidistant from the points B(3, p) and C(p, 5).
∴ AB = AC

\[\Rightarrow \sqrt{\left( 3 - 0 \right)^2 + \left( p - 2 \right)^2} = \sqrt{\left( p - 0 \right)^2 + \left( 5 - 2 \right)^2}\]                           (Distance formula)

Squaring on both sides, we get

\[9 + p^2 - 4p + 4 = p^2 + 9\]
\[ \Rightarrow - 4p + 4 = 0\]
\[ \Rightarrow p = 1\]

Thus, the value of p is 1.

 
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Chapter 6: Co-ordinate Geometry - Exercise 6.2 [Page 17]

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R.D. Sharma Mathematics [English] Class 10
Chapter 6 Co-ordinate Geometry
Exercise 6.2 | Q 37 | Page 17
R.D. Sharma Mathematics [English] Class 10
Chapter 6 Co-ordinate Geometry
Exercise 6.2 | Q 44 | Page 17

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Trigonometry in the form of triangulation forms the basis of navigation, whether it is by land, sea or air. GPS a radio navigation system helps to locate our position on earth with the help of satellites.
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  2. After 10 minutes, the guard observed that the boat was approaching the tower and its distance from tower is reduced by 240(`sqrt(3)` - 1) m. He immediately raised the alarm. What was the new angle of depression of the boat from the top of the observation tower?

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    Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

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