मराठी

A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.

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प्रश्न

A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.

बेरीज
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उत्तर

We know that the distance between the two points (x1​, y1​) and (x2​, y2​) is

d = `sqrt((x_2​−x_1​)^2+(y_2​−y_1​)^2​)`

Let the given points be A = (a, 7) and B = (−3, a) and the third point given is P(2, −1).

We first find the distance between P(2, −1) and A =(a, 7) as follows:

PA = `sqrt((x_2​−x_1​)^2+(y_2​−y_1​)^2)`

​= `sqrt((a−2)^2+(7−(−1))^2)`

​= `sqrt((a−2)^2+(7+1)^2)`

​= `sqrt((a−2)^2+8^2)​`

= `sqrt((a−2)^2+64)`​ 

Similarly, the distance between P(2,−1) and B = (−3, a) is:

PB = `sqrt((x_2​−x_1​)^2+(y_2​−y_1​)^2)`

= `sqrt((−3−2)^2+(a−(−1))^2)`

​= `sqrt((−5)^2+(a+1)^2)`

​= `sqrt(25+(a+1)^2)​` 

Since the point P(2,−1) is equidistant from the points A(a, 7) and B = (−3, a), therefore, PA = PB that is:

`sqrt((a−2)^2+64)​=sqrt(25+(a+1)^2)​`
⇒ (a − 2)2 + 64 = 25 + (a + 1)2

⇒ a2 − 4a + 4 + 64 = 25 + a2 + 2a + 1

⇒ a2 − 4a + 68 =   a2 + 2a + 26

⇒ −4a − 2a = 26 − 68

⇒ −6a = −42

⇒ a =`(−42)/(−6)` ​=7

Hence, a = 7.

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पाठ 28: Distance Formula - Exercise 28 [पृष्ठ ३३५]

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सेलिना Concise Mathematics [English] Class 9 ICSE
पाठ 28 Distance Formula
Exercise 28 | Q 7 | पृष्ठ ३३५

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Case Study

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.
A guard, stationed at the top of a 240 m tower, observed an unidentified boat coming towards it. A clinometer or inclinometer is an instrument used for measuring angles or slopes(tilt). The guard used the clinometer to measure the angle of depression of the boat coming towards the lighthouse and found it to be 30°.

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