Question

Find the point on the y – axis which is equidistant from the points
( 12, 3 ) and ( -5, 10 )

Answer 1

$\large\underline{\bold{Given \:Question - }}$

Find the point on the y – axis which is equidistant from the points ( 12, 3 ) and ( -5, 10 )

$\large\underline{\bold{ANSWER-}}$

Concept Used :-

Distance Formula :-

Let us assume a line segment joining the points A and B, then distance (D) between A and B is given by

$\rm D = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} }$

$\rm \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)$

Let's solve the problem now!!

• Let the coordinates of point on y - axis be P(0, y).

and

• Let the coordinates (12, 3) and (- 5, 10) are represented by A and B respectively.

Now,

• According to statement,

$\rm :\longmapsto\:PA = PB$

$\rm :\implies\:PA^{2} = PB^{2}$

$\rm :\longmapsto\: {(0 - 12)}^{2} + {(y - 3)}^{2} = {(0 + 5)}^{2} + {(y - 10)}^{2}$

$\rm :\longmapsto\: 144 + {y}^{2} + 9 - 6y = 25 + 100 + {y}^{2} - 20y$

$\rm :\longmapsto\:153 - 6y = 125 - 20y$

$\rm :\longmapsto\: - 6y + 20y = 125 - 153$

$\rm :\longmapsto\:14y = - 28$

$\rm :\longmapsto\:y = - 2$

$\bf\implies \:coordinate \: of \: point \: on \: y \: axis \: is \: (0, - 2)$

Additional Information :-

Section Formula :-

Let us assume a line segment joining the points A and B and let C divides AB in the ratio m : n internally, then coordinates of C is given by

$\boxed{ \sf \: (x, y) = (\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} )}$

$\rm \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)$

Midpoint Formula :-

Let us assume a line segment joining the points A and B and let C be the midpoint of line segment joinjng A and B, then coordinates of C is given by

$\boxed{ \sf \: (x, y) = (\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} )}$