How do you find a standard form equation for the line with P (5, 5), and Q (-3, -3)?

1 Answer
Apr 24, 2017

See the entire solution process below:

Explanation:

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)(-3) - color(blue)(5))/(color(red)(-3) - color(blue)(5)) = -8/-8 = 1#

Next, we can 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 first point from the problem gives:

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

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 now transform the equation we wrote to standard form as follows:

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

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

#-x + y - 0 = 0 - 0#

#-x + y = 0#

#color(red)(-1)(-x + y) = color(red)(-1) * 0#

#x - y = 0#

#color(red)(1)x + color(blue)(-1)y = color(green)(0)#