Question

Find the point on the X-asis which is equidistant from A-3, 4) and 1.-4)​

Answer 1

Given :

• A(-3, 4)
• B(1, -4)

To Find :

The point on the x-axis which is equidistant.

Solution :

Analysis :

Here the distance formula is used. First we have to find the distance from point A and then from B. After squaring both the distance we can get the point.

Required Formula :

$\\ :\boxed{\bf Distance\ Formula=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

where,

• (x₁, y₁) = Coordinates of point A
• (x₂, y₂) = Coordinates of point B

Explanation :

Let us assume that the point is P(x, 0) because the point is on x-axis.

• A(-3, 4)
• B(1, -4)

First PA :

• P(x, 0)
• A(-3, 4)

Using distance formula,

$\\ :\implies\sf Distance\ Formula=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

where,

• x₁ = x
• x₂ = -3
• y₁ = 0
• y₂ = 4

Using the required formula and substituting the required values,

$\\ :\implies\sf PA=\sqrt{(-3-x)^2+(4-0)^2}$

$\\ :\implies\sf PA=\sqrt{(-3-x)^2+(4)^2}$

$\\ :\implies\sf PA=\sqrt{(-3-x)^2+16}\qquad\Bigg\lgroup\bf eq.(i)\Bigg\rgroup$

Second PB :

• P(x, 0)
• B(1, -4)

Using distance formula,

$\\ :\implies\sf Distance\ Formula=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

where,

• x₁ = x
• x₂ = 1
• y₁ = 0
• y₂ = -4

Using the required formula and substituting the required values,

$\\ :\implies\sf PB=\sqrt{(1-x)^2+(-4-0)^2}$

$\\ :\implies\sf PB=\sqrt{(1-x)^2+(-4)^2}$

$\\ :\implies\sf PB=\sqrt{(1-x)^2+16}\qquad\Bigg\lgroup\bf eq.(ii)\Bigg\rgroup$

Now by squaring :

From eq.(i) and eq.(ii),

PA = PB,

$\\ :\implies\sf\sqrt{(-3-x)^2+16}=\sqrt{(1-x)^2+16}$

Now, PA² = PB²,

$\\ :\implies\sf(-3-x)^2+16=(1-x)^2+16$

Using the identity (a - b)² = a² - 2ab + b² in both LHS and RHS,

$\\ :\implies\sf(-3)^2-2.(-3).x+x^2+16=(1)^2-2.1.x+x^2+16$

$\\ :\implies\sf9-(-6x)+x^2+16=1-2x+x^2+16$

Cancelling 16 from both sides,

$\\ :\implies\sf9-(-6x)+x^2\ \cancel{+16}=1-2x+x^2\ \cancel{+16}$

$\\ :\implies\sf9+6x+x^2=1-2x+x^2$

Cancelling x² from both sides,

$\\ :\implies\sf9+6x\ \cancel{+x^2}=1-2x\ \cancel{+x^2}$

$\\ :\implies\sf9+6x=1-2x$

Transposing -2x to LHS and 9 to RHS,

$\\ :\implies\sf6x+2x=1-9$

After evaluation,

$\\ :\implies\sf8x=-8$

$\\ :\implies\sf x=\dfrac{-8}{8}$

$\\ :\implies\sf x=\cancel{\dfrac{-8}{8}}$

$\\ \therefore\boxed{\bf x=-1.}$

The point P(x, 0) = P(-1, 0)

The point is P(-1, 0).