First, we must 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)(-5) - color(blue)(-3))/(color(red)(1) - color(blue)(-3)) = (color(red)(-5) + color(blue)(3))/(color(red)(1) + color(blue)(3)) = -2/4 = -1/2#
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 values from the first point in the problem gives:
#(y - color(red)(-3)) = color(blue)(-1/2)(x - color(red)(-3))#
#(y + color(red)(3)) = color(blue)(-1/2)(x + color(red)(3))#
We can also substitute the slope we calculated and the values from the second point in the problem giving:
#(y - color(red)(-5)) = color(blue)(-1/2)(x - color(red)(1))#
#(y + color(red)(5)) = color(blue)(-1/2)(x - color(red)(1))#
We can solve for #y# to transform this equation to slope-intercept form. The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#
Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value.
#y + color(red)(5) = (color(blue)(-1/2) xx x) - (color(blue)(-1/2) xx color(red)(1))#
#y + color(red)(5) = -1/2x + 1/2#
#y + color(red)(5) - 5 = -1/2x + 1/2 - 5#
#y + 0 = -1/2x + 1/2 - (2/2 xx 5)#
#y = -1/2x + 1/2 - 10/2#
#y = color(red)(-1/2)x - color(blue)(9/2)#