First, we need to determine the slope of the line. The slope can be found by using the formula: #m = (color(red)(y_2) - color(blue)(y_1))/(color(red)(x_2) - color(blue)(x_1))#

Where #m# is the slope and (#color(blue)(x_1, y_1)#) and (#color(red)(x_2, y_2)#) are the two points on the line.

Substituting the values from the points in the problem gives:

#m = (color(red)(1/2) - color(blue)(7))/(color(red)(0) - color(blue)(-5)) = (color(red)(1/2) - color(blue)(7))/(color(red)(0) + color(blue)(5)) = (color(red)(1/2) - color(blue)(14/2))/5 = (-13/2)/(5/1) = -13/10#

The standard form of a linear equation is: #color(red)(A)x + color(blue)(B)y = color(green)(C)#

Where, if at all possible, #color(red)(A)#, #color(blue)(B)#, and #color(green)(C)#are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

The slope of an equation in standard form is: #m = -color(red)(A)/color(blue)(B)#

The #y#-intercept is of an equation in standard form is: #color(green)(C)/color(blue)(B)#

We know the slope so we can write:

#-13/10 = -color(red)(A)/color(blue)(B)#

Therefore, #color(red)(A) = 13# and #color(blue)(B) = 10#

From the point abobe we know the #y# intercept is #(0, 1/2)# and we know the value of #color(blue)(B)# so we can write and solve for #color(green)(C)#

#color(green)(C)/color(blue)(10) = 1/2#

#10 xx color(green)(C)/color(blue)(10) = 10 xx 1/2#

#color(blue)(cancel(color(black)(10))) xx color(green)(C)/cancel(color(blue)(10)) = 5#

#color(green)(C) = 5#

Substituting gives:

#color(red)(13)x + color(blue)(10)y = color(green)(5)#