Question

A point P on x-axis is equidistant from the point (7,6) and (-4,3).Find the co-ordinates of the point P.

Answer 1

@Michele_Laino

Answer 2

@amistre64

Answer 3

If I call with \((x,y)\) the coordinate of point \(P\), then I can write this:
\[\sqrt {{{\left( {x - 7} \right)}^2} + {{\left( {y - 6} \right)}^2}} = \sqrt {{{\left( {x - \left( { - 4} \right)} \right)}^2} + {{\left( {y - 3} \right)}^2}} \]

Answer 4

a line between 2 points, is full of points that are equidistant

Answer 5

P is on x-axis so y should be 0.

Answer 6

correct!
if we take the square of both sides, after a simplification I get:
\[{\left( {x - 7} \right)^2} + {\left( {0 - 6} \right)^2} = {\left( {x + 4} \right)^2} + {\left( {0 - 3} \right)^2}\]

Answer 7

pfft, thought it said P is a point ... well i spose 'on the x axis' narrows it down lol

Answer 8

x = 30/11?

Answer 9

correct!

Answer 10

Thanks.