Question

Find a point on the x-axis which is equidistant from point (7,6) and (-3,4).​

Answer 1

$\huge\fbox{\texttt{Answer = P(3,0)}}$

$\huge\rm{\green{ Explanation :-}}$

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Given :

The point ( say P ) lies on the x - axis. Therefore, y coordinate is 0.

Point P is equidistant from the points A(7,6) and B(-3,4).

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To Find :

Coordinates of the point P (on x - axis).

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Formula Used :

Distance Formula :-

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

where 'd' stands for distance.

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How to Find :

$\texttt{We\: are \:provided\: that\: the\: point \:P\: lies\: on \:x-axis. Therefore,}$

$\texttt{\green{coordinates\: of \:the\: point \:P = <strong>(</strong><strong>x</strong><strong>,</strong><strong>0</strong><strong>)</strong><strong>}</strong>}$

$\texttt{According \:to \:the \:question, }$

$\texttt{PA = PB (in\: terms \:of\: distance)}$

$\texttt{Also,}$ PA² = PB²

$\texttt{\red{Using <strong>distance</strong><strong> </strong><strong>formula</strong><strong>,</strong>}}$

$\implies (x-7)^2 + (-6)^2 = (x+3)^2 + (-4)^2$

$\implies x^2-14x + 49 + 36 = x^2 + 6x + 9 + 16$

$\implies -14x + 85 = 6x + 25$

$\implies 20x = 60$

$\texttt{Therefore,}$

$\huge\red{x = 3}$

$\texttt{\orange{Then,\: coordinates \:of\:point\:are :}}$

$\huge\fbox{\texttt{P(3,0)}}$