We have (1,3) is a midpoint between (x+1, 5) and (3, 2y)

The midpoint m(mx, my) :

mx = (xA+xB)/2

my =(yA + yB)/2

mx= 1 = (x+1 +3)/2

==> 2 = x+4

==> **x= -2**

my = 3 = (2y+5)/2

==> 2y + 5 = 6

==> 2y = 1

==> **y = 1/2**

Then we have the point: (-2+1, 5) = **(-1,5)**

and the point : (3, 2(1/2) = **(3,1)**

Where (1,3) is midpoint.

since (1,3) is the mid pt. of (x+1,5) and (3,2y)

therefore 1=(x+1+3)/2......i

and 3=(5+2y)/2......ii

on solving i and ii, we get,

1=(x+4)/2

therefore x=-2

6=5+2y

therefore y=0.5

hence the values of x and y are -2 and 0.5.

The mid point of the two points (x1,y1) and (x2,y2) is given by :

xM = (x1+x2)/2

yM = (y1+y2)/2

Here (x1,x2) = (x+1,5) and (x2,y2) = (3,3y) and (xM, yM) = (1,3) by hypothesis.

So XM = 1 = (x1+x2)/2 = {(x+1)+3}/2

1 = (x+1+3)/2

2 =x+4, x = 2-4 =-2

**x=-2**

yM =3 = (y1+y2)/2 = (5+2y)/2

3 = (5+2y)/2

6 = 5+2y , 2y = 6-5 ,** y = 1/2.**

So the solution for x appearing in x coordinate of first point is -2.

Solution for y appearing in y coordinate of the 2nd point is 1/2.

To calculate x and y, we'll use the formula of the midpoint of a line:

xM = (x+1+3)/2

yM = (5+2y)/2

But, the coordinates of xM and yM are given from enunciation, so:

xM = 1

1 = (x+1+3)/2

We'll cross multiply:

2 = x+4

We'll subtract 2:

x+2 = 0

**x = -2**

yM = 3

yM = (5+2y)/2

3 = (5+2y)/2

We'll cross multiply:

6 = 5+2y

6-5 = 2y

1 = 2y

We'll divide by 2:

y= 1/2

**y = 0.5**