Question

Write an equation of a line that is parallel to y=-6x+4 and passes through(2,10).

write an equation of line which is perpendicular to y=(-2/3)x-5 and passes through point (6,3).
write the equation of three lines so that the y-intercept of each is -3.

Answer 1

a)y=-6x+4
Slope and intercept a=-6 b=4
equation for new line:
y=cx+d
If new line is parallel c=a=-6 so
y=-6x+d
We substitude point (2,10)
10=-6*2+d /+12
d=22
so y=-6x+22
b)
y=-
`(2)/(3)`x-5
a=-2/3
b=-5
new line:
y=ex+f
If line is perpendicular e=-
`(1)/(a)`
so e=
`(3)/(2)`
we can write
y=
`(3)/(2)x+f`
Substituting point (6,3)
3=
`(3)/(2)*6+f`
f=3-9
f=-6
y=
`(3)/(2)x-6`