Question

Find the point on X-axis which is equidistant from (2,-5),and (-2,9)

Answer 1

Hello friend

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Let the point P(2,-5), Q(-2,9) and let be t the point on x-axis which will be R(t,0)

If R is an equidistant from P and Q,then we it can be written as

RP = RQ

So let's define it

By using the formula of equidistant

RP =
$\sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} }$
Here,

X2 = 2 Y2= -5

X1 = t Y1= 0

By putting the values we get,

RP =
$\sqrt{(2 - t) {}^{2} + ( - 5 - 0) {}^{2} } = \sqrt{(2 - t) {}^{2} + ( - 5) {}^{2} }$

RP =
$\sqrt{(2) {}^{2} + (t) {}^{2} - 2(2)(t) + 25} \\ \\ = \sqrt{4 + t { }^{2} - 4t + 25 } \\ \\ = \sqrt{t {}^{2} - 4t + 29}$

Similarly,we will prove another one,

RQ =
$\sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} }$
Here,

X2 = -2 Y2 = 9

X1 = t Y1 = 0

So,by putting the given values we get,

RQ =
$\sqrt{( - 2 - t) {}^{2} + (9 - 0) {}^{2} } \\ \\ = \sqrt{( - 2) {}^{2} + (t) {}^{2} + 2(2)(t) + (9) {}^{2} } \\ \\ = \sqrt{4 + t {}^{2} + 4t + 81} \\ \\ = \sqrt{t {}^{2} + 4t + 85}$

Now,we got the value of RP and RQ

We know that,

RP = RQ

So,let put the calculated values.

$\sqrt{t {}^{2} - 4t + 29} = \sqrt{t {}^{2} + 4t + 85} \\ \\ = ( \sqrt{t {}^{2} - 4t + 29 } ) {}^{2} = ( \sqrt{t {}^{2} + 4t + 85} ) {}^{2} \\ \\ = t {}^{2} - 4t + 29 = t {}^{2} + 4t + 85 \\ \\ substituting \: the \: vlue \: of \: right \: hand \: side \: to \: the \: left \: hand \: side \\ \\ = t {}^{2} - 4t + 29 - t {}^{2} - 4t - 85 = 0 \\ \\ = - 8t - 56 = 0 \\ \\ = - 8t = 56 \\ \\ t = \frac{56}{ - 8} \\ \\ t = - 7$

Therefore, the co-ordinat point on x-axis is (-7,0)

Thanks...

:)