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

Question

Julia Adams · Accepted Answer

#3x-2y=7# 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#

Genesis Allen · Answer

undefined To write the equation of a line in standard form that is perpendicular to $y = -\frac{2}{3}x + 4$ and passes through the point $(3, 1)$, first find the negative reciprocal of the slope of the given line. The negative reciprocal of $-\frac{2}{3}$ is $\frac{3}{2}$. Then use the point-slope form $y - y_1 = m(x - x_1)$ with the given point $(3, 1)$ and the slope $\frac{3}{2}$. Substitute these values into the equation to find the equation in point-slope form. After that, rearrange the equation into standard form $Ax + By = C$. Thus, the standard form of the equation is $3x - 2y = 7$.