Question

find the point on the x-axis which is equidistant from point (5, 4) and (-2,3)​

Answer 1

Answer:

$\boxed{\orange{\textsf{\textbf{ Required Point = (2,0) }}}}$

Step by step explanation:

We need to find out the point on x axis which os equidistant from the point (5,4) and (-2,3) . So that let us take that point be ( x , 0 ) .

Here y-coordinate will be 0 since the point lies on x axis .

Now here we can use the Distance Formula . As ,

$\boxed{\pink{\sf Distance =\sqrt{ (x_2-x_1)^2+(y_2-y_1)^2 }}}$

❒ Using this we have :-

$\tt : \implies D_1 = D_2 \\\\\\\tt : \implies \sqrt{ ( 5 - x )^2 + ( 4 - 0)^2} = \sqrt{ ( x +2)^2 + (3-0)^2 } \\\\\\\tt : \implies 25 + x^2 - 10x + 16 = x^2+4+4x + 9 \\\\\\\tt : \implies -10x -4x = 13-41 \\\\\\\tt : \implies -14x = -28 \\\\\\\tt : \implies \boxed{\pink{\tt x = 2 }}$

Hence the required answer is (2,0) .

Answer 2

Answer :-

(2, 0)

Topic:-

Co-ordinate Geometry

Given :-

• The point on the x - axis is equidistant from the point (5, 4) and (- 2, 3)

To find :-

The point

Understanding the Concept:-

Let the point be ( x , 0)  The y - co-ordinate is absent here as it on the x-axis So, As they given this point (x, 0) is equidistant from (5, 0) and (-2, 3) . Distance between them is equal . We find the distance between the points by distance formula that is

Required formula :-

$\red{\boxed{\maltese{Distance \:formula \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}}}$

Lets do !

Let the point (x,0) is P

(5, 4) is A

(-2,3) is B

Required condition:-

$\red{\boxed{\maltese{PA = PB}}}$

P = (x, 0) = (x_1 , y_1)

A = (5, 4) = (x_2, y_2)

P = ( x , 0) = (x_1, y_1)

B = (-2 ,3 ) = (x_2, y_2)

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

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

Squaring on both sides ,

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

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

Simplifying the equation ,

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

Transposing R.H.S equation to L.H.S

$(x-5)^2+16 -[(x+2)^2+9]=0$

$(x-5)^2+16 -(x+2)^2-9= 0$

$(x-5)^2 -(x+2)^2+7= 0$

$x^2-10x+25 -(x^2+4x+4)+7 =0$

$x^2-10x+25 -x^2-4x-4+7=0$

$\not{x^2}-10x+25 \not{-x^2}-4x-4+7=0$

$-14x+28=0$

$-14x = -28$

$\red{\boxed{x =2}}$

But According to our consideration that is (x,0) = (2,0)

$\red{\boxed{\maltese{The \:required\: point \:is\: (2, 0)}}}$