How do you write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (3,7) and perpendicular to 4x+5y=1?

1 Answer
Apr 21, 2018

Slope-intercept : #y = 5/4 x + 13/4#
Standard : #-5x + 4y = 13#

Explanation:

To write a linear equation in any form, you need at least two of the following three pieces of information: a point, a different point, and/or a slope. We are already given a single point, #(3, 7)#.

Note that if we have a line #y = mx + b# in slope-intercept form, its slope is given by #m#. The line perpendicular to this line will have a slope of #-1/m#. So let's put our provided equation in slope-intercept form.

#4x + 5y = 1#
#5y = 1 - 4x#
#y = 1/5 - 4/5 x = -4/5 x + 1/5#

The slope of the given line is then #-4/5#. The slope of the line perpendicular to this must be #5/4#. Now we have both a slope and a point.

To find our equation in slope-intercept form, we take the base form #y = mx + b# and plug in our slope of #m = 5/4# and our point of #(x, y) = (3, 7)# to find #b#.

#y = mx + b#
#y = 5/4 x + b#
#7 = 5/4 (3) + b#
#7 = 15/4 + b#
#7 - 15/4 = b#
#28/4 - 15/4 = b#
#13/4 = b#.

Having found #b#, we now know our equation in slope-intercept form. That is, #y = 5/4 x + 13/4#.

The standard form of a linear equation is #ax + by = c#, where #a, b, c# are constants. It is easy to move from slope-intercept form into standard form.

#y = 5/4 x + 13/4#
#4y = 5x + 13#
#-5x + 4y = 13#

Thus, our answer in standard form is #-5x + 4y = 13#.