How do you write an equation of a line that passes through points (0,5), (-3,5)?

The equation of a line can be written as

#y - y_0 = m * (x - x_0)#

where #(x_0, y_0)# is any point that lies on the line.

The gradient, #m#, of the line can be found using any two non-identical points that lie on the line.

# m = frac{y_2 - y_1}{x_2 - x_1}#

#= frac{5 - 5}{-3 - 0}#

#= 0#

A gradient of zero indicates that the line is horizontal.

The equation of the line can thus be simplified to

# y = y_0#

In this case, both points have a #y# coordinate of 5. The equation of this line is therefore

#y = 5#.

For this particular example, we could note that the value of the #y# coordinate is a constant: #5#, so the equation is:
#color(white)("XXX")y=5#

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For the general case:
#color(white)("XXX")#Given two points #color(red)(""(x_1,y_1)# and #color(blue)(""(x_2,y_2))#
#color(white)("XXX")#A two-point equation can be written as:
#color(white)("XXXXXX")(y-color(red)(y_1))/(x-color(red)(x_1))=(color(red)(y_1)-color(blue)(y_2))/(color(red)(x_1)-color(blue)(x_2))#

Substituting
#color(white)("XXX")color(red)(""(0,5))# for #color(red)(""(x_1,y_1))# and
#color(white)("XXX")color(blue)(""(-3,5))# for #color(blue)(""(x_2,y_2))#
we have:
#color(white)("XXX")(y-color(red)5)/(x-color(red)0)=(color(red)5-color(blue)(5))/(color(red)(0)-color(blue)(""(-3))#
Simplifying:
#color(white)("XXX")(y-5)/x=0#

#color(white)("XXX")y=5#