Let the point of x-axis be P(x, 0)Given A(2, -5) and B(-2, 9) are equidistant from PThat is PA = PBHence PA2\xa0= PB2 → (1)Distance between two points is\xa0{tex}\\sqrt{[(x_2\xa0- x_1)^2\xa0+\xa0(y_2\xa0- y_1)^2]}{/tex}PA =\xa0{tex}\\sqrt{[(2\xa0- x)^2\xa0+\xa0(-5\xa0- 0)^2]}{/tex}PA2\xa0= 4 – 4x +x2\xa0+ 25=\xa0x2\xa0- 4x + 29Similarly,\xa0PB2\xa0=\xa0x2\xa0+ 4x + 85Equation (1) becomesx2\xa0- 4x + 29 =\xa0x2\xa0+ 4x + 85- 8x = 56x = -7Hence the point on x-axis is (-7, 0)

X=8 chk your calculation
Let the point of x-axis be P(x, 0)Given A(2, -5) and B(-2, 9) are equidistant from PThat is PA = PBHence PA2 = PB2 \xa0→ (1)Distance between two points is\xa0√[(x2 – x1)2 +\xa0(y2\xa0- y1)2]PA =\xa0√[(2\xa0- x)2\xa0+\xa0(-5\xa0- 0)2]PA2\xa0= 4 – 4x +x2 + 25 =\xa0x2 – 4x + 29Similarly,\xa0PB2\xa0=\xa0x2\xa0+ 4x + 85Equation (1) becomesx2\xa0- 4x + 29 =\xa0x2\xa0+ 4x + 85- 8x = 56x = -7Hence the point on x-axis is (-7, 0)

We know that a point on the x-axis is of the form (x, 0). So, let the point P(x, 0) be equidistant from A(2, –5) and B(–2, 9). ThenPA = PB{tex}\\Rightarrow{/tex} PA2 = PB2{tex}\\Rightarrow{/tex} (2 – x)2 + (-5 – 0)2 = (-2 – x)2 + (9 – 0)2{tex}\\Rightarrow{/tex} 4 + x2 – 4x + 25 = 4 + x2 + 4x + 81{tex}\\Rightarrow{/tex} – 4x + 25 = 4x + 81{tex}\\Rightarrow{/tex} 8x = -56{tex}\\Rightarrow \\;x = \\frac{{ – 56}}{8} = – 7{/tex}Hence, the required point is (-7, 0)Check:{tex}PA = \\sqrt {{{\\{ 2 – ( – 7)\\} }^2} + {{( – 5 – 0)}^2}}{/tex}{tex}= \\sqrt {81 + 25} = \\sqrt {106}{/tex}{tex}PB = \\sqrt {{{\\{ – 2 – ( – 7)\\} }^2} + {{(9 – 0)}^2}}{/tex}{tex}= \\sqrt {25 – 81} = \\sqrt {106}{/tex}{tex}\\because{/tex} PA = PB{tex}\\therefore{/tex} Our solution is chekcked.

Now, the point is on x-axis, so the value \’y\’ of the given point will be zero.Let, the value of the \’x\’ value of the given point = x[Assume,x as a variable to do the further mathematical calculations.]So,the point is = (x,0)As mentioned in the question,the two points are equidistant from (x,0).So,Distance between (x,0) and (2,-5)= ✓(x-2)²+(0+5)²Distance between (x,0) and (-2,9).=✓(x+2)²+(0-9)²Now,the points are equidistant.So,✓(x-2)²+(0+5)² = ✓(x+2)²+(0-9)²(x-2)²+(0+5)² = (x+2)²+(0-9)²x²-4x+4+25 = x²+4x+4+81x²-4x-x²-4x = 4+81-4-25-8x = 56x = -7

Given points\xa0A(2,−5)\xa0and\xa0B(−2,9)Let the points be\xa0P(x,0).So,\xa0AP=PB\xa0and\xa0AP2=PB2\xa0⇒(x−2)2+(0+5)2=(x+2)2+(0−9)2⇒x2+4−4x+25=x2+4+4x+81⇒x2+29−4x=x2+85+4x⇒−4x−4x=85−29⇒−8x=56⇒x=−7Hence, point on the\xa0x-axis which is equidistant from\xa0(2,−5)\xa0and\xa0(−2,9)\xa0is\xa0(−7,0).

(0 , 2)
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2,-2