First, we must determine the slope. 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)(-3) - color(blue)(1))/(color(red)(3) - color(blue)(-5)) = (color(red)(-3) - color(blue)(1))/(color(red)(3) + color(blue)(5)) = -4/8 = -1/2#

We can now use the point slope formula to get an equation for the line. The point-slope formula states: #(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#

Where #color(blue)(m)# is the slope and #color(red)(((x_1, y_1)))# is a point the line passes through.

Substituting the slope we calculated and the the first point from the problem gives:

#(y - color(red)(1)) = color(blue)(-1/2)(x - color(red)(-5))#

#(y - color(red)(1)) = color(blue)(-1/2)(x + color(red)(5))#

We can now transform this equation into the standard form. 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

First, we will multiply each side of the equation by #2# to eliminate the fraction and allow each of the coefficients to be an integer.

#2(y - color(red)(1)) = 2 xx color(blue)(-1/2)(x + color(red)(5))#

#2y - 2 = color(blue)(cancel(color(black)(2))) xx color(blue)(-1/cancel(2))(x + color(red)(5))#

#2y - 2 = -(x + color(red)(5))#

#2y - 2 = -x - 5#

Next, add #color(red)(x)# and #color(blue)(2)# to each side of the equation to move both the #x# and #y# term to the left side of the equation and the constants to the right side of the equation:

#color(red)(x) + 2y - 2 + color(blue)(2) = color(red)(x) - x - 5 + color(blue)(2)#

#color(red)(x) + 2y - 0 = 0 - 3#

#color(red)(1)x + color(blue)(2)y = color(green)(-3)#