Question

Write the equation of the line passing through (3, 7) that is perpendicular to 5x − 6y = 4.

Answer 1

i got \[y=\frac{ 5 }{ 6 }x - \frac{ 27 }{ 6 }\]
but it is wrong

Answer 2

here were my steps:

to find m:
5x-6y=4
-6y=-5x+4
y=-5/6x - 4/6
m=5/6

line through (3,7) with slope m=5/6
y-y1=m(x-x1)
y-7=5/6(x-3)
y+7=5/6x-15/6
y=5/6x-15/6-42/6
y=5/6x-27/6

Answer 3

here is the example problem that i referenced

Answer 4

eq. of any line pependicular to ax+by=c is
bx-ay=d where d is a constant.
eq. of liine perpendicular to 5x-6y=4 is
6x+5y=c
\[\because~it ~passes~ through~ (3,7)\]6(3)+5(7)=c
c=18+35=53
so reqd. line is 6x+5y=53

Answer 5

ah thank you surj