Algebra linear equations city task (1)

Answers:

1) Graph Line 1

a) Graph: Please, see the attached graph.

b) (Line 1) Main Street (Equation in slope intercept form): y=(1/3)x+4  

2) Graph Line 2

a) Graph: Please, see the attached graph.

b) Y intercept: (0,10)  

c) X intercept: (5,0)

d) (Line 2) Street Equation in Standard Form: y+2x=10

3) Grap Line 3

a) Graph: Please, see the attached graph.

b) (Line 3) Street Equation in Point Slope Form: y-3=(5/3)(x+3)

Street Equation in Slope Intercept Form: y=(5/3)x+8

Solution:

1) Graph Line 1: (Main Street)

a) Graph the line formed from the two given points (9,7) (-9,1)

Please, see the attached graph.

b) Find the street equation in slope intercept form

P1=(9,7)=(x1,y1)→x1=9, y1=7

P2=(-9,1)=(x2,y2)→x2=-9,y2=1

m=(y2-y1)/(x2-x1)

m=(1-7)/(-9-9)

m=(-6)/(-18)

m=6/18

m=(6/6)/(18/6)

m=1/3

Point Slope form:

y-y1=m(x-x1)

y-7=(1/3)(x-9)

Slope Intercept Form:

y-7=(1/3)x-(1/3)(9)

y-7=(1/3)x-(1)(9)/3

y-7=(1/3)x-9/3

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

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

y=(1/3)x+4

2) Graph Line 2

a) Graph the line given the x and y intercepts

y intercept: 10

x intercept: 5

Please, see the attached graph

b) Y intercept  

y intercept: b=10

Y intercept: (0,b)→Y intercept: (0,10)

c) X intercept

x intercept: a=5

X intercept: (a,0)→X intercept: (5,0)

d) Find the Street Equation in Standard Form

P1=(0,10)=(x1,y1)→x1=0, y1=10

P2=(5,0)=(x2,y2)→x2=5, y2=0

m=(y2-y1)/(x2-x1)

m=(0-10)/(5-0)

m=(-10)/(5)

m=-2

Point Slope Form:

y-y1=m(x-x1)

y-10=(-2)(x-0)

y-10=(-2)(x)

y-10=-2x

Standard Form:

y-10+10+2x=-2x+10+2x

y+2x=10

3)  Graph Line 3

a) Graph the line given a point and the slope. (-3,3); m=5/3

Point 1: P1=(-3,3)=(x1,y1)→x1=-3, y1=3

Slope: m=5/3

m=dy/dx

5/3=dy/dx→dy=5, dx=3

Point 2: P2=(x2,y2)

x2=x1+dx→x2=-3+3→x2=0

y2=y1+dy→y2=3+5→y2=8

Point 2: P2=(x2,y2)→P2=(0,8)

Please, see the attached graph.

b) Find the Street Equation in Point Slope Form

Point Slope Form:

y-y1=m(x-x1)

y-3=(5/3)(x-(-3))

y-3=(5/3)(x+3)

Slope Intercept Form:

y-3=(5/3)x+(5/3)(3)

y-3=(5/3)x+(5)(3)/3

y-3=(5/3)x+15/3

y-3=(5/3)x+5

y-3+3=(5/3)x+5+3

y=(5/3)x+8

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap

1. sometimes

2. never true

3. sometimes

step-by-step explanation:

these are pretty self explanatory if you know what supp right and com angles are