Question

3. Find the point on the X - axis Which is equidistant from A (-3,4) and B (1,-4) .​

Answer 1

Explanation:

the coordinates of point which is present on x axis is (x, 0)

let P(x, 0) is equidistant point from A(-3, 4) and B(1, -4)

PA=PB

squaring on both sides

PA^2=PB^2

[x-(-3) ]+(0-4) =(x-1) +[0-(-4) ]

x+3) +16=x^2-2x+1+16

x^2+6x+9+16=x^2-2x+1+16

6x+9=-2x+1

6x+2x=9-1

8x=8

x=8/8

x=1

Answer 2

Given points:-

• A(-3, 4) and B(1, -4)

To find:-

• The point on x - axis which is equidistant from these points.

Solution:-

Let the point on x - axis be C(x, 0)

The point on x - axis is equidistant to points A and B

Hence, we can write,

• AC = BC ⟶ (i)

Let us calculate the distance between AC and BC separately by using distance formula.

For AC,

A(-3, 4) and C(x, 0)

Here,

x₁ = -3 and x₂ = x

y₁ = 4 and y₂ = 0

We know,

• Distance formula = √(x₂ - x₁) + (y₂ - y₁)

Putting the values in the formula,

AC = √[x - (-3)]² + [0 - 4]²

AC = √ (x + 3)² + (-4)²

Using identity (a + b)² = a² + 2ab + b²

AC = √x² + 6x + 9 + 16

AC = √x² + 6x + 25

Now,

For BC

B(1, -4) and C(x, 0)

x₁ = 1 and x₂ = x

y₁ = -4 and y₂ = 0

Putting the values in the distance formula,

BC = √[x - 1]² + [0 - (-4)]²

Using the identity (a - b)² = a² - 2ab + b²

BC = √x² - 2x + 1 + (4)²

BC = √x² - 2x + 1 + 16

BC = √x² - 2x + 17

Now,

From (i) we have AC = BC

Hence,

√x² + 6x + 25 = √x² - 2x + 17

Squaring on both sides,

= (√x² + 6x + 25)² = (√x² - 2x + 17)

⇒ x² + 6x + 25 = x² - 2x + 17

Bringing x² from RHS to LHS,

= x² - x² + 6x + 25 = -2x + 17

⇒ 6x + 25 = -2x + 17

Bringing all the variables on LHS and Constant on RHS,

= 6x + 2x = -25 + 17

⇒ 8x = -8

⇒ x = -8/8

⇒ x = -1

∴ The point on x - axis which is equidistant to points A(-3, 4) and B(1, -4) is (-1, 0).

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