Question

find the values of(x,y) if the distance of the point (x, y) from (-3,0) and (3,0) is 4​

Answer 1

Answer:

coordinates are (0,2.6)

Step-by-step explanation:

Let A's coordinate be(-3,0), B's coordinate be(3,0) and P's coordinate be (x,y)

    ∵, AP = 4 units

    ∴, 4²= $(\sqrt{(-3-x)^{2} +(0-y)^{2} })^{2}$

        16= 9 + x² + 6x + y²

          x²+y²=7-6x                 .....eq 1

  and,

      BP = 4 units

      4² = $(\sqrt{(3-x)^{2}+(0-y)^{2} )}^{2}$    

      16= 9 + x² - 6x + y²

      x²+y²=7+6x                   ......eq 2

equating eq 1 and eq 2, we get :-

      7-6x=7+6x

        0=12x

        x=0

putting value of x in eq 1

     0+ y²=7+0

       y=√7

       y=2.6

∴, the coordinates of P are (0,2.6)

Answer 2

Given:-

1. The distance of the point (x, y) from (-3,0) and (3,0) is 4.

To find:-

• Find the values of (x,y)..?

Solutions:-

The distance d between two points (x1, y1) and (x2, y2) is given by the formula.

• d = √(x1 - x2)² + (y1 - y2)²

(x, y) is equidistant from both (-3, 0) and (3, 0).

• Let d1 be the distance between (x, y) and (-3, 0)
• Let d2 be the distance between (x, y) and (3, 0)

So, using the distance formula for both these pair of point. We have,

=> d1 = √(x + 3)² + (y - 0)²

=> d2 = √(x - 3)² + (y - 0)²

Now, since both these distance are given to be the same.

Let the equate both d1 and d2

=> √(x + 3)² + (y - 0)² = √(x - 3)² + (y - 0)²

Squaring on both sides. We have,

=> (x + 3)² + (y - 0)² = (x - 3)² + (y - 0)²

=> x² + 9 + 6x + y² = x² + 9 - 6x + y²

=> 12x = 0

=> x = 0

Also,

The value of both d1 and d2 is 4 unit.

Substituting the value of x in the equation d1.

=> d1 = √(x + 3)² + (y - 0)²

=> 4 = √(0 + 3)² + (y - 0)²

=> 4 = √9 + y²

Squaring on both sides of the equation. We have,

=> 16 = 9 + y²

=> y² = 16 - 9

=> y² = +,- 7

Hence, the value of x and y are x = 0 , y = +,- 7