Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4).
Answer:-
Let the point on the x-axis be P(a, 0).
\[PA = \sqrt{\left( a - \left( - 3 \right) \right)^2 + \left( 0 - 4 \right)^2}\]
\[ = \sqrt{\left( a + 3 \right)^2 + 16}\]
\[PB = \sqrt{\left( a - 1 \right)^2 + \left( 0 - \left( - 4 \right) \right)^2}\]
\[ = \sqrt{\left( a - 1 \right)^2 + 16}\]
\[ {PA}^2 = {PB}^2 \]
\[ \Rightarrow \left( a + 3 \right)^2 + 16 = \left( a - 1 \right)^2 + 16\]
\[ \Rightarrow a^2 + 6a + 9 = a^2 - 2a + 1\]
\[ \Rightarrow 8a = - 8\]
\[ \Rightarrow a = - 1\]
\[\left( a, 0 \right) = \left( - 1, 0 \right)\]
Chapter 5. Co-ordinate Geometry – Practice Set 5.1 (Page 107)