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)(0) - color(blue)(-10))/(color(red)(-4) - color(blue)(0)) = (color(red)(0) + color(blue)(10))/(color(red)(-4) - color(blue)(0)) = 10/-4 = -5/2#
Because the first point in the problem has a value of #0# for #x# we know this is the #y#-intercept. The y-intercept is therefore: #color(blue)(-10)#
We can now write and equation in 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.
Substituting the slope we calculated and the #y#-intercept we determined gives:
#y = color(red)(-5/2)x + color(blue)(-10)#
#y = color(red)(-5/2)x - color(blue)(10)#
We need to convert this equation to the Standard Form for Linear equations. 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 can add #5/2x# to each side of the equation to put the #x# and #y# variables on the left side of the equation as required by the Standard formula:
#5/2x + y = 5/2x + color(red)(-5/2)x - color(blue)(10)#
#5/2x + y = 0 - color(blue)(10)#
#5/2x + y = -10#
We can now multiply each side of the equation by #color(red)(2)# to eliminate the fraction as required by the Standard formula:
#color(red)(2)(5/2x + y) = color(red)(2) xx -10#
#(color(red)(2) xx 5/2x) + (color(red)(2) xx y) = -20#
#(cancel(color(red)(2)) xx 5/color(red)(cancel(color(black)(2)))x) + 2y = -20#
#color(red)(5)x + color(blue)(2)y = color(green)(-20)#
