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Straight Lines — Applied Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceApplied MathsStraight Lines1 Mark
MCQ
A point on the $X$-axis, which is equidistant from the point $(7,6)$ and $(3,4)$ is
  • A
    $\left(0, \frac{15}{2}\right)$
  • ✓
    $\left(\frac{15}{2}, 0\right)$
  • C
    $\left(\frac{2}{15}, 0\right)$
  • D
    $\left(\frac{8}{15}, 0\right)$
✓

Answer

Correct option: B.
$\left(\frac{15}{2}, 0\right)$
(B) $\left(\frac{15}{2}, 0\right)$
Explanation : Let $P(x, 0)$ be the point, which is equidistant from $A(7,6)$ and $B(3,4)$
$\therefore$ $P A=P B$
$\Rightarrow$ $P A^2=P B^2$
$\Rightarrow$ $(7-x)^2+(6-0)^2=(3-x)^2+(4-0)^2$
$\Rightarrow$ $49-14 x+x^2+36=9-6 x+x^2+16$
$\Rightarrow$ $x=\frac{60}{8}=\frac{15}{2}$
$\therefore$ The point on $X$-axis is $\left(\frac{15}{2}, 0\right)$.

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