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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.

Sagot :

Luanaidn
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=-[tex] \frac{2}{3} [/tex]x-5
a=-2/3
b=-5
new line:
y=ex+f
If line is perpendicular e=-[tex] \frac{1}{a} [/tex]
so e=[tex] \frac{3}{2} [/tex]
we can write 
y=[tex] \frac{3}{2}x+f [/tex]
Substituting point (6,3)
3=[tex] \frac{3}{2}*6+f [/tex]
f=3-9
f=-6
y=[tex] \frac{3}{2}x-6 [/tex]
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