Question

Q1. Find a point on the x-axis is equidistant from the points (7,6) and (-3,4)​

Answer 1

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}

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

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

(7−x)

2

+(6−0)

2

=

(−3−x)

2

+(4−0)

2

⇒ 49 + x^2 -14x + 36 = 9 + x^2 +6x + 1649+x

2

−14x+36=9+x

2

+6x+16

⇒85 -14x = 25 +6x85−14x=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)

Answer 2

Step-by-step explanation:

Plz refer to attachment for the answer..

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