find the point on x axis which are equidistant from points p (-3,4)&q(1,-4)​

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1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Let the point on x – axis be (x, 0) which is equidistant from the points A (- 3, 4) and B (1, – 4).

   We know,

   Distance between two points A and B is given by

   [tex]\bf\implies \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }[/tex]

   [tex]\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)[/tex]

   Now,

   According to statement,

   [tex]\rm :\longmapsto\:PA = PB[/tex]

   On squaring both sides, we get

   [tex]\rm :\longmapsto\:PA^{2} = PB^{2} [/tex]

   [tex]\rm :\longmapsto\: {(x + 3)}^{2} + {(0- 4)}^{2} = {(x – 1)}^{2} + {(0 + 4)}^{2} [/tex]

   [tex]\rm :\longmapsto\: {x}^{2} + 9 + 6x + 16 = {x}^{2} + 1 – 2x + 16 [/tex]

   [tex]\rm :\longmapsto\:9 + 6x= 1 – 2x[/tex]

   [tex]\rm :\longmapsto\:6x + 2x= 1 -9[/tex]

   [tex]\rm :\longmapsto\:8x = – 8[/tex]

   [tex]\bf\implies \:x = – 1[/tex]

   Hence,

   [tex]\bf:\longmapsto\:The \: point \: on \: x – axis \: be \: P \: (-1,0)[/tex]

   Additional Information :-

   Section Formula

   Section Formula is used to find the co ordinates of the point(Q) Which divides the line segment joining the points (B) and (C) internally in the ratio m : n

   [tex]{\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}[/tex]

   Midpoint Formula :-

   Midpoint Formula is used to find the midpoint of line segment joining two points

   [tex]{\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{x_2 + x_1}{2}, \dfrac{y_2 + y_1}{2}\Bigg) \quad}}}}[/tex]

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