Question

the point P on x axis equity distance from the points A (- 1, 0) close and B (5,0) close is​

Answer 1

• Point P on x-axis = P ( x , 0 )
• A ( -1 , 0 )
• B ( 5 , 0 )

All coordinate lies in x-axis so we can find coordinate of P by mid- point formula.

$</strong><strong> </strong><strong>\</strong><strong>b</strong><strong>f</strong><strong> </strong><strong>x</strong><strong> = \frac{{x}_{1} + {x}_{2}}{2} \\ \\ </strong><strong>\</strong><strong>bf </strong><strong>x</strong><strong> = \frac{ - 1 + 5}{2} \\ \\ \large\underline{\boxed { \bf \orange{x = 2}}}$

Coordinates of point p ( 2 , 0 )

Answer 2

The required point is P (2,0)

Step-by-step explanation:

the given points be A(-1,0) and B (5,0)

let the required point be P (x,0)

given that points are equidistant

therefore ,

(PA)² = (PB)²

(-1-x)²+(0-0)² = (5-x)²+(0-0)²

(-1)² +(-x)²+2(-1)(-x) = 5² +x²-10x +0

1+x²+2x=5²+x²-10x

12x= 24

x= 2

hence The required point is (2,0)

#Learn more:

Find the distance of point P(-3,-2) and (5,0) from x-axis and y-axis​

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