Question

find coordinates of point on x axis which is equidistant from point (7,6) and (-3,4)
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Answer 1

Step-by-step explanation:

7 and -3 is the coordinates of the point on x- axis

Answer 2

Given

Two points A(7,6) & B(-3,4)

To Find :-

We have to find the point in x-axis

Solution:-

Let the required point be P(x,0)

• As we know that on x-axis y=0

Now using Distance formula :-

$\implies\sf\ \ D= \sqrt{\big(x_2-x_1\big)^2+\big(y_2-y_1\big)^2}$

So , as they are equidistant

$\therefore\sf \ AP= PB\\ \\ : \implies\sf\ AP^2=PB^2\\ \\ \\ : \implies\sf \bigg\{\sqrt{\big(x-7\big)^2+\big(0-6\big)^2}\bigg\}^2=\bigg\{\sqrt{\big(-3-x\big)^2+\big(4-0\big)^2}\bigg\}^2\\ \\ \\ :\implies\sf\ \big(x-7\big)^2+\big(-6\big)^2=\big(-3-x\big)^2+\big(4\big)^2\\ \\ \\ :\implies\sf\ \cancel{x^2}+49-14x+36= \cancel{x^2}+9+6x+16\\ \\ \\ :\implies\sf\ 85-14x= 25+6x\\ \\ \\ \\ :\implies\sf\ 85-25= 6x+14x\\ \\ \\ \\ :\implies\sf\ 60=20x\\ \\ \\ :\implies\sf\ x= \cancel{\dfrac{60}{20}}\\ \\ \\ :\implies\underline{\boxed{\red{\sf\ x= 3}}}$

$\sf\underline{\therefore\ The\ Required\ point \ be \ \big(3,0\big)}$