Question

72. The point on the y-axis which is equidistant from A(-5,-2) and B(3, 2) is :
A) (1,-4)
B) (2,0)
C) (0, -3)
D) (-4,0)

please answer​

Answer 1

Answer:

P ( 0, - 2 ) are the coordinates of the points which are equidistant from the given points.

NOTE: Answer is not there in the given options.

Step-by-step-explanation:

We have given that,

A ≡ ( - 5, - 2 ) ≡ ( x₁, y₁ )

B ≡ ( 3, 2 ) ≡ ( x₂, y₂ )

Let P be the point on Y-axis which is equidistant from points A and B.

∴ PA = PB - - ( 1 )

P lies on Y-axis.

∴ Its x coordinate is 0.

Let P ( 0, y )

Now,

PA = PB - - [ From ( 1 ) ]

√[ ( x - x₁ )² + ( y - y₁ )² ] = √[ ( x - x₂ )² + ( y - y₂ )² ]

⇒ √{ [ 0 - ( - 5 ) ]² + [ y - ( - 2 ) ]² } = √[ ( 0 - 3 )² + ( y - 2 )² ]

⇒ √[ ( 0 + 5 )² + ( y + 2 )² ] = √[ ( - 3 )² + ( y - 2 )² ]

By taking squares of both sides, we get,

⇒ ( 0 + 5 )² + ( y + 2 )² = ( - 3 )² + ( y - 2 )²

⇒ ( 5 )² + ( y )² + 2 × y × 2 + ( 2 )² = 9 + ( y )² - 2 × y × 2 + ( 2 )²

⇒ 25 + y² + 4y + 4 = 9 + y² - 4y + 4

⇒ 25 + 4y = 9 - 4y - - [ Cancelling y² & 4 ]

⇒ 25 - 9 = - 4y - 4y

⇒ 16 = - 8y

⇒ y = - 16 / 8

⇒ y = - 2

∴ y - coordinate of point P is - 2.

∴ P ( 0, - 2 ) are the coordinates of the points which are equidistant from the given points.

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Additional Information:

1. Distance Formula:

The formula which is used to find the distance between two points using their coordinates is called distance formula.

d ( A, B ) = √[ ( x₁ - x₂ )² + ( y₁ - y₂ )² ]

2. Section Formula:

The formula which is used to find the coordinates of a point which divides a line segment in a particular ratio is called section formula.

• x = ( mx₂ + nx₁ ) / ( m + n )

• y = ( my₂ + ny₁ ) / ( m + n )

Answer 2

Let ,

The point on the y axis be " (0,y) "

Given ,

• The point on the y-axis is equidistant from A(-5,-2) and B(3, 2)

We know that , the distance between two points is given by

$\boxed{ \sf{D = \sqrt{ {( x_{2} - x_{1} )}^{2} + {(y_{2} - y_{1} )}^{2} } }}$

Thus ,

$\sf \mapsto \sqrt{ { \{0 - ( - 5) \}}^{2} + { \{y - ( - 2) \}}^{2} } = \sqrt{ { \{0 - 3 \}}^{2} + { \{y - 2 \}}^{2} } \\ \\ \sf \mapsto squaring \: on \: both \: sides \: we \: , \: get \\ \\ \sf \mapsto 25 + {(y)}^{2} + 4 + 4y = 9 + {(y)}^{2} + 4 - 4y \\ \\ \sf \mapsto 16 = - 8y \\ \\ \sf \mapsto y = - \frac{16}{8} \\ \\ \sf \mapsto y = - 2$

Therefore ,

• The point on the y axis is (0,-2)