Question

the distance of the point (a;b) from (-3;0) and (3;0)are 4 units each then the value of a;b are

Answer 1

Answer:

the answer is (0,+ -√7)

Answer 2

Answer:

(a,b) = (0,√7),(0,-√7)

Step-by-step explanation:

Given,

The distance of the point (a,b) from (-3,0) and (3,0) are 4 units

To find,

The value of a and b

Recall the formula

Distance formula: Distance AB between two points A(x₁, y₁ ) and B(x₂, y₂) is given by

AB = $\sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}$

Solution:

Let O be the point (a,b), A be the point (-3,0) and B be the point (3,0)

It is given that OA = OB = 4 units

Distance between the points O(a,b) and A(-3,0) = OA

= $\sqrt{((-3) - a)^2 + (0-b)^2}$

= $\sqrt{9 + a^2 + 6a + b^2}$

Since OA = 4 units, we have

 $\sqrt{9 + a^2 + 6a + b^2}$  = 4

Squaring on both sides,

9 + a² + 6a +b² = 16

a² + 6a +b² = 16 - 9 = 7

a² + 6a +b²  = 7 -----------------------(1)

Distance between the points O(a,b) and B(3,0) = OB

= $\sqrt{(3 - a)^2 + (0-b)^2}$

= $\sqrt{9 + a^2 - 6a + b^2}$

Since OA = 4 units, we have

 $\sqrt{9 + a^2 - 6a + b^2}$  = 4

Squaring on both sides,

9 + a² - 6a +b² = 16

a² - 6a +b² = 16 - 9 = 7

a² - 6a +b²  = 7 -----------------------(2)

Subtracting (2) from (1) we get

12a = 0

a = 0

Substituting a= 0, equation (1) we get

b²  = 7

b = ±√7

The point (a,b) = (0,√7),(0,-√7)

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