Question

Find a point on x-axis which s equidstant from A(5,4) and B(-2,3)

Answer 1

Hey there!

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Q. Find the point on the x - axis which is equidistant from (5, 4) and (-2, 3).

Solution :

Since the point on x - axis have its ordinate = 0.

So, P(x, 0) is any point on the x - axis.

Since P(x, 0) is equidistant from A(5, 4) and B(-2, 3).

PA = PB ⇒ $PA^{2} = PB^{2}$

⇒ $(x - 5)^{2} + (0 - 4)^{2} = (x + 2)^{2} + (0 - 3)^{2}$

⇒  $x^{2}$ + 25 - 10x + 16 = $x^{2}$ + 4 + 4x +9

⇒ $x^{2}$ + 41 - 10x = $x^{2}$ + 13 + 4x

⇒ -10x - 4x = 13 - 41

⇒ - 14x = - 28

⇒ $x = \frac{-28}{-14}$ = 2

∴ The point equidistant from given points on the x - axis is (2, 0).

Answer 2

A point on the x-axis will have a coordinate with y = 0

Define x:

Let x be the value of the x-coordinate.

⇒ s = (x , 0)

Form equation and solve for x:

Distance between A and S must be equal to B and S

$\sqrt{(X_{A1} - X_{S1} )^2 + (Y_{A1} - Y_{S1} ) ^2 }  = \sqrt{(X_{B1} - X_{S1} )^2 + (Y_{B1} - Y_{S1} ) ^2 }$

Given that A(5, 4),  B(-2, 3) and S(x, 0):

$\sqrt{(5 - x )^2 + (4-0) ^2 }  = \sqrt{(-2-x)^2 + (3-0 ) ^2 }$

Evaluate the constants:

$\sqrt{(5 - x )^2 + 16 }  = \sqrt{(-2-x)^2 + 9 }$

Square both sides:

$(5 - x )^2 + 16 = (-2-x)^2 + 9$

Evaluate ( a ± b)² = a² ± 2ab + b²:

$25 - 10x + x^2 + 16 = 4 +4x + x^2 + 9$

Subtract x² from both sides:

$41 - 10x = 13 +4x$

Moved all x to LHS and constant on RHS:

$14x = 28$

Divide both sides by 14:

$x = 2$

Find the coordinate s:

s = ( 2, 0)

Answer: The point s is (2, 0)