Find the point / \( \mathrm{s} \) on the \( x \)-axis at the distance 13 from the point \( (11,12) \).

Given: 

Point $( 11,\ 12)$.

To do: 

We have to find the point/s  on the \( x \)-axis at the distance 13 from the point \( (11,12) \).

Solution: 

Let point $A=( 11,\ 12)$

Co-ordinates of point on $x-axis\ B=( x,\ 0)$

According to the question,

$\Rightarrow AB=\sqrt{( x-11)^2+( 0-12)^2}$

$\Rightarrow 13=\sqrt{(x-11)^2+144}$

Squaring on both sides, we get,

$169=(x-11)^2+144$

$(x-11)^2=169-144$

$(x-11)^2=25$

$x-11=\sqrt{25}$

$x-11=\pm 5$

$x=5+11$ or $x=-5+11$

$x=16$ or $x=6$

Therefore, the required points are $(6, 0)$ and $(16,0)$. 

Tutorialspoint

Updated on: 10-Oct-2022

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