Question
Answers

Find a point on x-axis which is equidistant from points (5, 4) and (-2, 3).
\[
  {\text{A}}{\text{. }}\left( {2,0} \right) \\
  {\text{B}}{\text{. }}\left( { - 2,0} \right) \\
  {\text{C}}{\text{. }}\left( {3,0} \right) \\
  {\text{D}}{\text{. }}\left( { - 3,0} \right) \\
\]

Answer Verified Verified
Hint: Calculate the individual distance from each of the given points to the point on x-axis, both these distances must be equal. Use the formula for distance between two points. Also, the Y-coordinate for any point which lies on x-axis is zero.

Complete step-by-step answer:
Given Data –
Points are (5, 4) and (-2, 3).
Let the point on x-axis be (x, 0).
The formula for calculating the distance (d) between two points $\left( {{{\text{x}}_1},{{\text{y}}_1}} \right)$ and $\left( {{{\text{x}}_{2,}}{{\text{y}}_2}} \right)$ respectively is
${\text{d = }}\sqrt {{{{\text{(}}{{\text{x}}_1}{\text{ - }}{{\text{x}}_2}{\text{)}}}^2} + {{({y_1} - {y_2})}^2}} $.
Let the distance from point (5, 4) to (x, 0) be D1.
\[{\text{D1 = }}\sqrt {{{(5 - x)}^2} + {{(4 - 0)}^2}} \] -> Equation 1.
Let the distance from point (-2, 3) to (x, 0) be D2.
${\text{D2 = }}\sqrt {{{( - 2 - x)}^2} + {{(3 - 0)}^2}} $ -> Equation 2.
Equidistant implies both the distances D1 and D2 are equal;
⟹ Equation 1 = Equation 2
⟹ $\sqrt {{{(5 - {\text{x)}}}^2} + {{(4 - 0)}^2}} = \sqrt {{{( - 2 - {\text{x)}}}^2} + {{(3 - 0)}^2}} $
Squaring on both sides
⟹ \[{\left( {{\text{5 - x}}} \right)^2} + {\left( 4 \right)^2} = {\left( { - 2 - {\text{x}}} \right)^2} + {\left( 3 \right)^2}\]
⟹ ${\text{25 + }}{{\text{x}}^2} - 10{\text{x + 16 = 4 + }}{{\text{x}}^2} + 4{\text{x + 9}}$
⟹ 28 = 14x
⟹ x = 2.
Hence the point is (2, 0) which is Option A.
Note –
In problems like this always find the distance from each of the given points respectively and then equate the distance equations obtained to determine the value of the required point.
Distance from (5, 4) – D1 and (-2, 3) – D2 are obtained and equated and solved in order to determine the point on x-axis which is (2, 0).



Bookmark added to your notes.
View Notes
×