Question

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

Answer 1

$\mathtt{\bf{\small{\underline {\red{Required\: answer \: :- }}}}}$

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

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 ,

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

Hence the required answer is (2,0) .

                    ____________

Answer 2

$\huge\boxed{\fcolorbox{white}{pink}{AnsweR}}$

Since the required point (say P) is on the X-axis, its ordinate will be zero. Let the abscissa of the point be x. Therefore, coordinates of the point P are (x, 0). Let A and B denote the points (5, 4) and (-2, 3), respectively. Since we are given that AP = BP, we have AP2 = BP2 i.e., (x - 5)2 + (0 - 4)2 = (x + 2)2 + (0 - 3)2 or x2 + 25 - 10x + 16 = x2 + 4 + 4x + 9 or -14x = - 28 or x = 2 Thus, the required point is (2, 0).

                    ____________