Question

Find a point on the y-axis which is equidistant from the points A (6,5) and B (- 4,3)

Answer 1

Hint: In this question remember that distance between the two points is given as; $XY = \sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $ here X is starting point whose coordinate is (a, b) and Y is the end point whose coordinate is (c, d), use this information to approach the solution.

Complete step-by-step solution:
Given points are A (6,5) and B (- 4,3)
Let us suppose the point on the y-axis which is equidistant from the given points be P (0, y) because as it lies on the y-axis so its x - coordinate will be 0

For point P to be equidistant from points A and B, distance between points A and P will be equal to the distance between points B and P
AP = BP (equation 1)
According to distance formula, the distance between any two points X (a, b) and Y (c, d) is given by
$XY = \sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $
Now substituting the values in the above equation, we get
$AP = \sqrt {{{\left( {0 - 6} \right)}^2} + {{\left( {y - 5} \right)}^2}} = \sqrt {36 + {{\left( {y - 5} \right)}^2}} $
And $BP = \sqrt {{{\left( {0 - \left( { - 4} \right)} \right)}^2} + {{\left( {y - 3} \right)}^2}} = \sqrt {16 + {{\left( {y - 3} \right)}^2}} $
Now substituting the values of AP and BP in equation 1 we get
$\sqrt {36 + {{\left( {y - 5} \right)}^2}} = \sqrt {16 + {{\left( {y - 3} \right)}^2}} $
Squaring the above equation both sides, we have
$36 + {\left( {y - 5} \right)^2} = 16 + {\left( {y - 3} \right)^2}$
By applying the algebraic formula ${\left( {a \pm b} \right)^2} = {a^2} + {b^2} \pm 2ab$in the above equation we get
$36 + {y^2} + 25 - 10y = 16 + {y^2} + 9 - 6y$
$ \Rightarrow $$36 + 25 - 10y + 6y - 9 - 16 = {y^2} - {y^2}$
$ \Rightarrow $$36 - 4y = 0$
$ \Rightarrow $$4y = 36$
$ \Rightarrow $y = 9
Therefore, the point on the y - axis which is equidistant from the points A (6, 5) and B (- 4, 3) is P (0, 9).

Note: In these types of problems, the statement of the problem should be carefully converted into an equation, which will be used to equate the given data and the unknown parameter which we need to find out. Here, if instead of the y-axis there would have been an x-axis so the coordinates of the point assumed would be (x, 0) instead of (0, y).