How do you write the standard form of a line given (7,2) and (3, -5)?

1 Answer
Jul 21, 2017

See a solution process below:

Explanation:

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)(-5) - color(blue)(2))/(color(red)(3) - color(blue)(7)) = (-7)/-4 = 7/4#

We can now use the point-slope formula to write 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 values from the first point in the problem gives:

#(y - color(red)(2)) = color(blue)(7/4)(x - color(red)(7))#

We can now transform this equation to the Standard Form for a Linear Equation. 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

#y - color(red)(2) = (color(blue)(7/4) xx x) - (color(blue)(7/4) xx color(red)(7))#

#y - color(red)(2) = 7/4x - 49/4#

#y - color(red)(2) + 2 = 7/4x - 49/4 + 2#

#y - 0 = 7/4x - 49/4 + (4/4 xx 2)#

#y = 7/4x - 49/4 + 8/4#

#y = 7/4x - 41/4#

#-color(red)(7/4x) + y = -color(red)(7/4x) + 7/4x - 41/4#

#-7/4x + y = 0 - 41/4#

#-7/4x + y = -41/4#

#color(red)(-4)(-7/4x + y) = color(red)(-4) xx -41/4#

#(color(red)(-4) xx -7/4x) + (color(red)(-4) xx y) = -cancel(color(red)(4)) xx -41/color(red)(cancel(color(black)(4)))#

#(-cancel(color(red)(-4)) xx -7/color(red)(cancel(color(black)(4)))x) - 4y = 41#

#color(red)(7)x - color(blue)(4)y = color(green)(41)#