Math, asked by nairdevika26, 8 months ago

5. Find the co-ordinates of a point P on y-axis so
that PA = PB where A = (-2, 4) and B = (-5, -3).​

Answers

Answered by Anonymous
21

Given:-

PA = PB

A = (-2, 4) = (x1, y1) and B = ( -5, -3) = (x2, y2)

To Find:-

co-ordinates of point P on y axis.

Solution:-

Point p is on y axis.

•°• coordinate of point P should be 0 and y i.e (0, y)

°•° PA = PB •°• PA² = PB²

  • By Distance formula

(0 - (-2))² + (y - 4)² = (0 - (-5))² + (y - (-3))²

(0 + 2)² + (y - 4)² = (0 + 5)² + (y + 3)²

2² + (y² - 8y + 16) = 5² + (y² - 6y + 9)

4 + y² - 8y + 16 = 25 + y² - 6y + 9

- 8y + 20 = - 6y + 34

- 8y + 20 = -6y + 34

- 8y + 6y = 34 - 20

- 2y = 14

y = 14/-2 = -7

•°• y = -7

Answer:- The co-ordinate of point P on y axis is 0 and -7, i.e. P = (0, -7).

Answered by Anonymous
4

Given ,

  • A = (-2,4) and B = (-5,-3)

  • The coordinate of point P is on the y axis

  • The distance between point P and A is equal to the point P and B

Let , The coordinate of P be " (0 , y)

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}  } }}

According to the question ,

 \sf \mapsto \sqrt{ {(0 -   \{- 2 \})}^{2}  +  {(y - 4)}^{2} }  =  \sqrt{ {(0 -  \{- 5 \})}^{2}  +  {(y -  \{- 3 \})}^{2} }  \\  \\  \sf{squaring \:  on  \: both \:  sides \:  , \:  we \:  get  } \\  \\ \sf \mapsto  4 +  {(y)}^{2}  + 16 - 8y = 25 +  {(y)}^{2}  + 9 +6y \\  \\ \sf \mapsto  20  - 8y = 34 + 6y \\  \\ \sf \mapsto   - 7y = 14 \\  \\ \sf \mapsto  y =  - 2

Therefore ,

The coordinate of point P is (0 , -7)

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