The equation of the line is:

y-y1= m(x-x1)

Let us use one of the point (2,3)

==> y-3= m(x-2)

==> y= m(x-2)=3

Now let us calculate the slope (m):

m= y2-y1/(x2-x1)= 3-3/ -3-2= 0/-5=0

Then the slope m= 0

==> **y= 3**

**That means that we have a horizontal line parallel to x axis and passes through y=3**

Coordinates : (2,3) , (-3,3) , (x,y)

(3-3)/(-3-2) = (y-3)/[x-(-3)]

0 = (y-3)/(x+3)

0 = y-3

**y = 3**

A line which passes through 2 given points, A(2,3) and B(-3,3), has the following equation:

(xB-xA)/(x-xA) = (yB-yA)/(y-yA)

(-3-2)/(x-2) = (3-3)/(y-3)

-5/(x-2) = 0/(y-3)

We'll cross multiply:

-5(y-3) = 0(x-2)

-5y+15 = 0

We'll isolate -5y to the left side. For this reason, we'll subtract 20 both sides:

-5y = -15

We'll divide by -5:

y = 3

The equation of the line is **y=3.**

The general equation of a line is:

y = mx + c

Where:

m = slope of the line, and

c is the y coordinate of line where it intercepts the x-axis.

The slope of a line passing through any line passing through 2 points (x1, y,1) and (x2, y2) is given by:

Slope = (y2 -y1)/(x2 - x1)

Therefore:

Slope of line passing through the given points = m = (3 - 3)/-3 -2) = 0/-5 = 0

Thus the line is parallel to x-axis, and the value of m in equation of the line is 0.

As the y-coordinates of both the given points are 3, it is clear that this line intercepts y-axis at point (0, 3).

Therefore the value of c in equation of line is 3.

Substituting these values of m and c in the standard equation of line we get equation of line as:

y = 0*x + 3 = y

Answer:

Equation of line is:

y = 3

Let us suppose that a line y = mx +c passes through the points (2,3) and (-3,3). Now our work is to find m and c by the conditions that the points (2,3) and (3, -3) satisfy the equation y = mx+c, or mx+c = y.

Since (2,3) satisfy mx+c = y, put x=2 and y = 3.

m(2)+c = 3.

2m+c = 3..................(1)

Similarly, the other point (3,3) also should satisfy mx+c = y. So,

3m+c = 3..................(2)

Now we solve for m and n from equations (1) and (2):

Eq(3)- Eq(1) eliminates c.

(3m+c)-(2m+c) = 3-3

m = 0.

Substituting m =0 in (1), c = 3.

So y= mx+c now becomes, y = 0x+3 . Or y =3 is the equation that passes through (2,3) and (3,3)..