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Straight Lines — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSStraight Lines2 Marks
Question
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
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Answer

Let P(x, 0) be any point on the x-axis which is equidistant from Q(7, 6) and R (3, 4).
Then $PQ = \sqrt {{{(x - 7)}^2} + {{(0 - 6)}^2}}$ $= \sqrt {{x^2} - 14x + 49 + 36}$
$ = \sqrt {{x^2} - 14x + 85}$
$PR = \sqrt {{{(x - 3)}^2} + {{(0 - 4)}^2}} = \sqrt {{x^2} - 6x + 9 + 16}$
$= \sqrt {{x^2} - 6x + 25}$
Since PQ = PR
$\therefore \sqrt {{x^2} - 14x + 85} = \sqrt {{x^2} - 6x + 25}$
Squaring both sides, we have
x2 - 14x + 85 = x2 - 6x + 25
$\Rightarrow$ -14x + 6x = 25 - 85 $\Rightarrow$ 8x = -60
$\Rightarrow x = \frac{{15}}{2}$
Thus coordinates of point on the x-axis is $\left( {\frac{{15}}{2},0} \right)$.

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