Question

find a point on X axis which is equestrian from A(2,-5) B(-2,9)
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Answer 1

Assumption: let the required point be (x,0). This point is equidistant from points A(2,—5) and B(—2,9). This also means that the distances between the required point and these 2 points individually are equal.

Formula To Be Applied: The Distance Formula

$\\\qquad\bullet\qquad\underline{\boxed{\sf Distance_{required}= \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1}} )^{2}}}\bigstar$

Solution:

$\\\quad\longrightarrow\quad\sf AC = BC$

$\\\quad\longrightarrow\quad\sf (x-2)^{2} + (0+5)^{2} = (x+2)^{2}+(0-9)^{2}$

$\\\quad\longrightarrow\quad\sf x^{2} +4-4x + 25 = x^{2} +4+4x + 81$

$\\\quad\longrightarrow\quad\sf -8x = 56$

$\\\quad\therefore\quad\sf x = -7$

Hence, the required point is (-7,0)

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Answer 2

Let the point of x-axis be P(x,0)

Given A(2,−5) and B(−2,9) are equidistant from P

That is, PA=PB

Hence PA square=PB square  ---(1)

Distance between two points is square root of [(x square −x1)the whole square+(y square −y1)the whole square] ------distance formula

PA= Square root of [(2−x)2+(−5−0)2]

PA square=4−4x+x square+25=x square −4x+29

and PB= Square root of [(−2−x) square+(9−y) square]

PB square=x square+4x+85

Equation (1) becomes

x square −4x+29=x square+4x+85

−8x=56

x=−7

Hence the point on x-axis is (−7,0)