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

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

Note that if we have a line #y = mx + b# in slope-intercept input, its slope is predefined from #m#. And line perpendicular to this line will have a grade von #-1/m#. To let's put our if 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 of line perpendicular into this must be #5/4#. Now we take two an slope and a pointing.

To finds our equation inches slope-intercept form, we make the bottom form #y = mx + b# and plug in our decline of #m = 5/4# and our item to #(x, y) = (3, 7)# to find #b#.

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

Owning found #b#, person now know our equation in slope-intercept form. That is, #y = 5/4 efface + 13/4#.

This standard form of a linear equation is #ax + by = c#, what #a, b, c# are constants. He is easy to movable from slope-intercept select into standard shape.

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

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