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[30 POINTS] Please help!!!

[30 POINTS] Please help!!!-example-1
User CBlew
by
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1 Answer

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Answer:

Part 1)
y=1.5x+5

Part 2)
y=-(2/3)x-(11/3)

Part 3)
y=0.25x+2.75

Part 4)
y=-2x+5

Part 5)
y=0.5x-1

Part 6) The graph in the attached figure

Explanation:

Part 1) we have


m=3/2=1.5


point(-2,2)

The equation of the line into point slope form is equal to


y-y1=m(x-x1)

substitute


y-2=1.5(x+2)


y=1.5x+3+2


y=1.5x+5

Part 2) we know that

If two lines are perpendicular

then

the product of their slopes is equal to minus one

so


m1*m2=-1

the slope of the line 1 is equal to


m1=1.5

Find the slope m2


1.5*m2=-1


m2=-2/3

Find the equation of the line 2

we have


m2=-2/3


point(-7,1)

The equation of the line into point slope form is equal to


y-y1=m(x-x1)

substitute


y-1=(-2/3)(x+7)


y=-(2/3)x-(14/3)+1


y=-(2/3)x-(11/3)

Part 3) we have


m=1/4=0.25


point(1,3)

The equation of the line into point slope form is equal to


y-y1=m(x-x1)

substitute


y-3=0.25(x-1)


y=0.25x-0.25+3


y=0.25x+2.75

Part 4) we have


m=-2


b=5 -----> y-intercept

we know that

The equation of the line into slope intercept form is equal to


y=mx+b

substitute the values


y=-2x+5

Part 5) we have that

The slope of the line 4 is equal to
-2

so

the slope of the line perpendicular to the line 4 is equal to


-2*m=-1\\m=(1/2)=0.5

therefore

in this problem we have


m=0.5


point(-2,-2)

The equation of the line into point slope form is equal to


y-y1=m(x-x1)

substitute


y+2=0.5(x+2)


y=0.5x+1-2


y=0.5x-1

Part 6)

using a graphing tool

see the attached figure

[30 POINTS] Please help!!!-example-1
User Anil Bhomi
by
7.5k points

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