Answer:
Step-by-step explanation:
eq of line perpendicular to 3x+5y=10 is
5x-3y=a
where a is constant.
it passes through (3,2)
5(3)-3(2)=a
a=15-6=9
reqd. eq. is 5x-3y=9
or
3y=5x-9
y=5/3 x-3
gradient=5/3
y-intercept=-3
or
5y=-3x-10
y=-3/5 x-2
slope=-3/5
slope of reqd. line=-1/(-3/5)=5/3
eq. of line through (3,2) is
y-2=5/3(x-3)
3y-6=5x-15
3y=5x-15+6
3y=5x-9
y=5/3 x-3
