Math, asked by reddymanojkumar74, 9 months ago

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

Answers

Answered by Mounikamaddula

Answer:let the point on x-axis is p(x,0)

From the above question,

It is equidistant from the points,

A(7,6), B(-3,4)

It means PA=PB

√(7-x)²+(6-0)²=√(-3-x)²+(4²)

√49+x²-14x+36=√x²+9+6x+16

√x²-14x+85=√x²+6x+25

-14x+85=6x+25

20x=60

x=3

So the point is p(3,0)...

Step-by-step explanation:

Hope it helps you frnd........

Answered by VelvetBlush

105

$\huge\star\underline\mathfrak\red{Answer:-}$ $$

➡Let the point on X axis be P(x, 0) which is equidistant from the points A(7,6) and B(–3,4)

➡That means, PA = PB

➡Using the distance between two points formula $\sqrt{(x_{2} - x_{1})^2+(y_{2} - y_{1})^2}$

⇒ $\sqrt{(7 - x)^2+(6 - 0)^2}$ = $\sqrt{(-3 - x)^2+(4 - 0)^2}$

⇒ 49 + x^2 -14x + 36 = 9 + x^2 +6x + 16

⇒ 85 -14x = 25 +6x

⇒ 20x = 60

⇒x = 3

The point on X axis which is equidistant from the points (7,6) and (–3,4) is (3,0)

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