How do you write the standard form of the equation given (3,1) and slope 1/2?

1 Answer
Feb 26, 2017

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

Or

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

Explanation:

First, we can use the point-slope formula to find an equation for this 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 values from the problem gives:

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

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

We can transform our equation into the standard form as follows:

First, multiply each side of the equation by #color(red)(2)# to eliminate all fractions while keeping the equation balanced:

#color(red)(2)(y - color(red)(1)) = color(red)(2) xx color(blue)(1/2)(x - color(red)(3))#

#(color(red)(2) xx y) - (color(red)(2) xx color(red)(1)) = x - color(red)(3)#

#2y - 2 = x - 3#

Next, subtract #color(red)(x)# and add #color(blue)(2)# to each side of the equation to put the #x# and #y# terms on the left side of the equation while keeping the equation balanced:

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

#-x + 2y - 0 = 0 - 1#

#-x + 2y = -1#

Now, multiply each side of the equation by #color(red)(-1)# to convert the #x# coefficient to a positive integer while keeping the equation balanced:

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

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

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