Question

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

Answer 1

Answer:

We know that a point on the y-axis is of the form (0,y).So,let the point p(0,y) be equidistant from A and B .Then

$= > (6 - 0)^{2} +( 5 - y)^{2} = ( - 4 - 0)^{2} + ( {3 - y)}^{2} \\ \\ = > 36 + 25 + {y}^{2} - 10y = 16 + 9 + {y}^{2} - 6y \\ \\ = > 4y = 36 \\ \\ = > y =9$

so the required point is (0,9)

Let us check our solution.....

$ap = \sqrt{ {(6 - 0)}^{2} + {(5 - 9)}^{2} } \\ \\ \: \: \: = \sqrt{36 + 16} \\ \\ \: \: = \sqrt{52}$

$bp = \sqrt{ {( - 4 - 0)}^{2} + (3 - 9)^{2} } \\ \\ \: \: \: \: = \sqrt{16 + 32} \\ \\ \: \: \: = \sqrt{52}$

we see that (0,9) is the intersection of the y- axis and the perpendicular bisector of ab.

Answer 2

Step-by-step explanation:

We Have :-

$point \: a \: (6,5) \\ \\ \\ point \: b \: ( - 4,3)$

To Find :-

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

Method Used :-

$distance \: formula$

Solution :-

$we \: know \: that \: any \: point \: on \: y - axis \: is \: of \: type \: (0, y) \\ \\ \\ let \: this \: point \: be \: c(0, y) \\ \\ \\ then \: distance \\ \\ \\ ac = bc \\ \\ \\ using \: distance \: formula \\ \\ \\ \sqrt{ac} = \sqrt{bc} \\ \\ squaring \: both \: sides \\ \\ \\ ac = bc \\ \\ \\ (0 - 6)^{2} + (y - 5) ^{2} = (0 + 4)^{2} (y - 3)^{2} \\ \\ \\ 36 + (y^{2} + 25 - 10y) = 16 + (y^{2} + 9 - 6y) \\ \\ \\ 36 + y^{2} + 25 - 10y= 16 + y^{2} + 9 - 6y \\ \\ \\ 61 + y^{2} - 10y = 25 + y^{2} - 6y \\ \\ \\ 61 - 10y = 25 - 6y \\ \\ \\ 61 - 25 = - 6y + 10y \\ \\ \\ 36 = 4y \\ \\ \\ y = \dfrac{36}{4} \\ \\ \\ y = 9 \\ \\ \\ so \: the \: point \: is \: (0,9)$