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)