Question

find a point on x axis ,which is equidistant from the points (7,6) and (3,4)​

Answer 1

Step-by-step explanation:

Let the point be P

let coordinates of P be (x,0)

since the point is equidistant from. A(7,6) and B(3,4)

AP=BP

use distance formula

AP=√{(x2-x1)^2+(y2-y1)^2}

on solving

AP=√(x^2+49-14x+36)

AP=√(x^2-14x+85)

BP=√{(x3-x1)^2+(y3-y1)^2}

on solving

BP=√(9+x^2-6x+16)

=√(x^2-6x+25)

AP=BP

√(x^2-14x+85)=√(x^2-6x+25)

squaring both sides

x^2-14x+85=x^2-6x+25

60=8x

x=7.5

thus P=(7.5,0)

Answer 2

Answer:

(3,0)

Step-by-step explanation:

Points are A(7,6) and B(-3,4) .

On x - axis, y = 0.

Apply Distance formula:

⇒ (x - 7)^2 + (0 - 6)^2 = (x + 3)^2 + (0 - 4)^2

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

=> x^2 + 49 - 14x + 36 - x^2 - 9 - 6x - 16 = 0

=>  60 - 20x = 0

=> x = 3

Hence, the coordinates are (3,0)

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