Question

By plotting the point P(5,-3) and Q(4,6) , find by counting methods the point on y-axis which is equidistant from P and Q.​

Answer 1

Answer:

idk

Step-by-step explanation:

I need the answer too

Answer 2

Answer:

( 0, 1 )

Step-by-step explanation:

Midpoint of a segment with endpoints at ( $x_{1}$ , $y_{1}$ ) and ( $x_{2}$ , $y_{2}$ ) is a point with coordinates ( $\frac{x_{1} +x_{2} }{2}$ , $\frac{y_{1} +y_{2} }{2}$ )

Intersection of perpendicular bisector to segment PQ and y-axis is equidistant from P and Q.

$m_{PQ}$ = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$ = $\frac{-3-6}{5-4}$ = - 9

Slope of perpendicular line to PQ is opposite reciprocal to $m_{PQ}$ and equal to $\frac{1}{9}$

Coordinates of midpoint of PQ are ( 4.5, 1.5 )

Equation of the perpendicular bisector is

y - 1.5 = $\frac{1}{9}$ ( x - 4.5 )

y = $\frac{1}{9}$ x + 1 and x = 0

( 0, 1 )