Question

Find the coordinates of points on the x- axis which are at a distance of 17 units from the point \[\left( {11, - 8} \right)\].

Answer 1

Hint: In this question, we simply have to find the point on the x axis when we know another point in the same plane (the given point is situated in the 4th quadrant), and the distance between two points is also mentioned as 17 units.
Formula used: \[c = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]

Complete step-by-step answer:
Here, in this question, we are given one point, that is, \[\left( {11, - 8} \right)\] and another point on the same plane is to be calculated, also the distance between two points is mentioned.
According to the question, the points to be calculated lie on the x- axis only. So, the position on the x axis is assumed to be \[\left( {x,0} \right)\].
Also, distance of \[\left( {x,0} \right)\] from \[\left( {11, - 8} \right) = 17\]
So, by using the above formula, and substituting the coordinates respectively, we get:
\[ \Rightarrow \sqrt {{{\left( {x - 11} \right)}^2} + {{\left( { - 8 - 0} \right)}^2}} = 17\]
Now, on further solving the question by squaring both the sides, the root on R.H.S gets removed, and we get:
\[ \Rightarrow {\left( {x - 11} \right)^2} + {\left( 8 \right)^2} = {\left( {17} \right)^2}\]
\[
   \Rightarrow {\left( {x - 11} \right)^2} = 289 - 64 \\
   \Rightarrow {\left( {x - 11} \right)^2} = 225 \;
 \]
Now, on further expanding \[{\left( {x - 11} \right)^2}\], by using the identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], we get:
\[
   \Rightarrow {x^2} + 121 - 22x = 225 \\
   \Rightarrow {x^2} - 22x + 121 - 225 = 0 \\
   \Rightarrow {x^2} - 22x - 104 = 0 \;
 \]
By splitting 22x as follows, and further solving the question, we get:
\[
   \Rightarrow {x^2} - 26x + 4x - 104 = 0 \\
   \Rightarrow x\left( {x - 26} \right) + 4\left( {x - 26} \right) = 0 \;
  \]
Therefore, \[x = \left( {26, - 4} \right)\]
Hence, the coordinates on the plane would be \[x = \left( {26,0} \right)or\,\left( { - 4,0} \right)\]

Note: The question itself mentioned that the point is present on the x- axis only, that is the y coordinate of that point should be taken as 0. Therefore, the points that we got represent x axis only, two answers are possible here.