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Straight Lines — Maths STD 11 Science — Question

CBSE 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
$x^2 - 14x + 85 = x^2 - 6x + 25$
$\Rightarrow  -14x + 6x = 25 - 85  \Rightarrow 8x = -60$
$\Rightarrow x = \frac{{15}}{2}$
Thus coordinates of point on the x-axis is $\left( {\frac{{15}}{2},0} \right)$.

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