How do you write the standard form of the equation through: (3,1), perpendicular to y=-2/3x+4?

1 Answer
Nov 6, 2016

Answer:

#3x-2y=7#

Explanation:

Write the standard form of the line that goes through #(3,1)# and is perpendicular to #y=-2/3 x +4#.

The equation #y=color(red)(-2/3) x +4# is in slope intercept form
#y=color(red)mx+b# where #color(red)m#= slope and #b# = the #y# intercept.

The slope of this line is then #m=color(red) (-2/3)#

A perpendicular slope is the opposite sign reciprocal. So, we change the sign of #color(red)(-2/3)# and switch the numerator and denominator.

Perpendicular slope #color(blue)m= color(blue)(3/2)#

To find the equation of the new line, use the point slope equation
#y-y_1=m(x-x_1)# where #m=# slope and #(x_1, y_1)# is a point.

The slope is #color(blue)(3/2)# and the point is the given point #(3,1)#.

#y-1=color(blue)(3/2)(x-3)color(white)(aaa)#Distribute

#y-1=3/2 x -9/2#

Standard form is #ax+by=c# where #a, b and c# are integers and #a# is positive.

#color(white)(aa)2(y-1=3/2x -9/2)color(white)(aaa)#Multiply the equation by #2#

#color(white)(aaaaa)2y-2=3x-9#
#-3xcolor(white)(aaaaaaa)-3xcolor(white)(aaa)#Subtract #3x# from both sides

#-3x+2y-2=-9#
#color(white)(aaaaaaaa)+2color(white)(aaa)+2color(white)(aaa)#Add #2# to both sides

#-3x+2y=-7#

#-1(-3x+2y=-7)color(white)(aaa)#Multiply the equation by #-1#

#3x-2y=7#