How do you find the equation of the line that passes through (5, -3), in Standard Form, and is parallel to the line that passes through (4, -5) and (-1, -6)?

1 Answer
Stefan V.
Aug 30, 2015

#-x + 5y = -20#

Explanation:

You know that you must find the equation of a line that passes through point #(5,-3)#, and that is parralel to another line that passes through two points, #(4,-5)# and #(-1,-6)#.

The important thing to realize here is that two parralel lines have the same slope, but different #y#-intercepts.

This means that you can use the two points given for the second line to determine its slope, which will be equal to the slope of the target line.

For a general form line that passes through two points #(x_1,y_1)# and #(x_2,y_2)#, the slope is equal to

#color(blue)("slope" = m = (y_2 - y_1)/(x_2 - x_1))#

The slope of the two lines will be

#m = (-6 - (-5))/(-1 - 4) = ((-1))/((-5)) = 1/5#

You now have the slope of the target line, which means that you can use the point #(5, -3)# to write its equation in point slope form

#color(blue)(y - y_1 = m * (x - x_1))#

#y - (-3) = 1/5 * (x - 5)#

#y + 3 = 1/5(x-5)#

The standard form of a line is

#color(blue)(Ax + By = C)#

Rearrange your equation to get

#y = 1/5x - 1 - 3#

#-1/5x + y = -4" "# #<=># #" "-x + 5y = -20#

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