Geometry Honors 07-08 1st Semester

Thursday, September 20, 2007

#4.5 Writing Equations of Lines

Aim: to write equations of lines
  • (1, 2, 5-8)
  • (9-11)
  • (13)
  • (15 - 34) all

We would like to thank Algebra and Geometry for inviting us to their wedding. We wish them nothing but love and matrimonial bliss for eternity!

I will offer 3 extra credit points to the FIRST student who answers AND thoroughly explains a problem correctly! The maximum # of extra credit points you can earn is 3 points. That means, once you've received extra credit for a problem, you can't do another one to get 6 extra credit points.

Also, once the problem has been confirmed (by me) that it has been correctly solved and explained, it is no longer eligible for extra credit. For example, if #30 has already been posted by someone, but you don't agree with his/her answer, post your solution/explanation. The first person may be wrong. Again, you won't know until I confirm. Please make sure your posts contain:

  • the question:
  • For example, #30, find the equation of the line containing median WO if W(0,0) H(8, 3) and Y(2, 9)
  • your last name and period
  • correct answer
  • explanation

There are 34 questions in all and 53 students in all. That means all but 19 of you can receive extra credit. Don't procrastinate! The early bird gets the worm! Get your extra credit NOW! Good Luck!

P.S. don't forget that your center of mass triangle on cardstock with string is due tomorrow!!!!

posted by msichan at 9/20/2007 12:30:00 PM

35 Comments:

  • maria per-4

    #1
    y=4x-3

    find the slope and the y-int

    m= 4 y-int=-3

    ]is the line that seperates x and y

    x]y (x,y chart)
    ----
    1]1
    2]5
    3]9

    now graph the points
    connect points w/ straight edge

    By Blogger *maria*, At September 20, 2007 at 3:28 PM  

  • 4.5 #2
    Adam Shemony - Period 3
    Graph the Linear Equations: y = -2x +1

    find the slope and y intercept of the equation

    slope: -2
    y intercept: (0,1)
    ^point which the line crosses the y-axis

    x]y <--- (x,y) chart

    1]-1
    2]-3
    3]-5

    Now that you have some of the points such as (0,1) (1,-1) (2,-3) (3,-5)
    you can graph them and connect them with a ruler or straightedge

    By Blogger shemony, At September 20, 2007 at 3:42 PM  

  • This comment has been removed by the author.

    By Blogger HEATHER!, At September 20, 2007 at 3:45 PM  

  • laura per-4
    #6
    y=6

    since there is no x,the slope of this equation is zero so it must be vertical.

    make a graph
    mark the y int (6)
    make a vertical

    (laura's internet wasn't working so she has to use my name)

    By Blogger JGH, At September 20, 2007 at 3:57 PM  

  • #15

    Determine whether each pair of lines are parallel, perpendicular, or neither. y = 4x + 5, y = 4x - 3.

    Pena P.3

    parallel

    The reason the pair of lines are parallel is because the two line have congruent slopes (4) and the both have congruent complimentary angles (75 degrees).

    By Blogger HEATHER!, At September 20, 2007 at 4:05 PM  

  • This comment has been removed by the author.

    By Blogger JGH, At September 20, 2007 at 4:06 PM  

  • #3

    Q- graph the linear equation
    y = -3x - 5

    m = 3 y-int = 0, -5

    You have to find the slope and the y intersept first.
    To do this, you need to make a chart an X and Y as shown where y is the the equation y = -3x-5

    X ___ Y
    -2 ___ 1
    -1 ___ -2
    0 ___ -5
    1 ___ -8

    The x and the y of the chart are your cooridinates (ex. -2,1 and 0,-5).Once you plug your coordinates in your graph, connect the points with a straightedge.

    Cilia Nazef
    Period 3

    By Blogger C.nazef, At September 20, 2007 at 4:11 PM  

  • 4.5 #16
    y=-4x-3

    because the slopes are not the same the lines do intersect.
    -4x-3=-4x-3

    is how you write the problem to solve for x and find the intersection
    But they have the same y-intercept which is -3
    -3 is their point of intersection or the y-intercept

    The slopes are negative but the lines are not perpendicular to each other. They do not make a 90 degree angle.So , they are also not negative recipricals.

    Stern P.4
    :] hi ms.chan

    By Blogger sabby-stern, At September 20, 2007 at 4:34 PM  

  • This comment has been removed by the author.

    By Blogger ladrido.c.4, At September 20, 2007 at 4:37 PM  

  • NOW this is Hesse from Period 4.

    #7

    2x=y-3

    put in slope int. form

    add 3 to each side)

    you get- y=2x+3

    make an x,y chart

    x|y
    -----
    1|5
    2|7
    3|9

    graph the points

    connect the points using a straight edge.

    By Blogger JGH, At September 20, 2007 at 4:39 PM  

  • #9. for each line, determine the equation in slope-intercept form

    point A (0,2)
    point B (4,-2)
    m = y1-y2= 2+2= 4= -1
    x1-x2= 0-4= -4

    y = -1(0)+2


    im confused if we distribute after
    since -1 x 0 = 0.
    or if we leave it like that because slope intercept equation
    is

    y=mx+b.

    :]

    By Blogger ladrido.c.4, At September 20, 2007 at 4:39 PM  

  • This comment has been removed by the author.

    By Blogger Guido, At September 20, 2007 at 4:54 PM  

  • Brian per.3

    #8
    Graph the Following linear equations
    2y-3x=12

    Add 3x from both sides to get:
    2y=3x+12
    Then didve 2 from both sides to get:y=3/2x+6



    Slope=3/2 Y-int=(0,6)
    After that make a x/y chart starting with the y intercept at 0 and 6, then plug the numbers as they go on in the equation 3/2x+6.
    X---Y
    o---6
    1---7 1/2
    2---9
    3---10 1/2

    The points are (0,6),(1,7.5),(2,9)
    (3,10.5)

    Graph them and connect the dots with a ruler or straightedge

    By Blogger BrianC., At September 20, 2007 at 4:57 PM  

  • Blackwell per.4

    #18

    Determine whether each pair of lines is parallel, perpendicular, or niether. If they intersect, find their point of intersection.

    y=(3/2)x + 8
    y=(-2/3)x -5

    The two lines are perpendicular because their slopes are opposite reciprocals.

    Their point of intersection is
    (-6,-1)

    By Blogger J.Blackwell, At September 20, 2007 at 5:12 PM  

  • #17
    You convert the two formulas, and make them into a point slope form. Then you get, 4/3x - 5 = 3/4x + 2.
    Once you solve this, you can get the slope and you find that the lines' slopes are reciprocals but not opposite. Which means that they are not perpendicular.And they are not parallel because the slopes are not the same. Which brings you to neither

    By Blogger dani.herrera.p.4, At September 20, 2007 at 5:34 PM  

  • 4.5 #16
    The problem is
    y=-4x-3
    y= 4x-3

    By Blogger sabby-stern, At September 20, 2007 at 5:35 PM  

  • 4.5 #17
    Determine whether each pair of lines are parallel,perpendicular, or niether. If they intersect find their point of intersection.

    y=4/3x-5 and y=3/4x+2

    To see if lines are parallel you look at the slope, and determine if they have the same number.
    To see if lines are perpendicular you look at the slope, and determine if they are opposite recipricols.
    In this case the slope is not the same. They are recipricols, however NOT opposite.
    In that case the answer is niether.

    By Blogger cgomez, At September 20, 2007 at 5:37 PM  

  • Vanessa Pulgar *P:3

    # 10
    Determine the equation in slope-intercept form.

    Point 1: (-5, -8)
    Point 2: (3, -5)

    First, find the slope
    Then, plug in slope (m) and b (y intercept)

    y=mx+b
    m= Y2 - Y1 = (-5) - (-8)
    X2 - X1 = 3 - (5-)

    m= 3/8
    b= -8

    y= 3/8x-8

    By Blogger Unknown, At September 20, 2007 at 5:39 PM  

  • This comment has been removed by the author.

    By Blogger fell in a well, At September 20, 2007 at 6:02 PM  

  • rodriguez p.4

    #19

    determine whether or not each set of points is collinear

    (1,2)(3,4)(5,6)
    plot the points

    or you can do
    Y2-Y1
    X2-X1

    4-2 = 2 reduced = 1
    3-1 = 2

    6-4 = 2 reduced = 1
    5-3 = 2

    since they hav the same slope they are collinear

    By Blogger fell in a well, At September 20, 2007 at 6:03 PM  

  • This comment has been removed by a blog administrator.

    By Anonymous Anonymous, At September 20, 2007 at 6:36 PM  

  • Fung-On. Period 3

    4.5 # 23

    Find the equation, in slope- intercept form , of the line through each of the following pairs of points.

    (1,2) (3,4)

    find the slope of the line between points (1,2) and (3,4)
    By dividing the difference between the y-coordinates and x-coordinates:
    2-4/1-3 = -2/-2 = 1
    therefore the slope is 1

    Find the slope intercept form using either points equaling the slope of 1:

    y-2/x-1 = 1
    y-2 = x-1
    solve for y
    therefore y=x+1

    answer: y=x+1

    By Anonymous Anonymous, At September 20, 2007 at 6:36 PM  

  • This comment has been removed by the author.

    By Blogger Mkaladi, At September 20, 2007 at 6:42 PM  

  • Kaladi period 3

    24. Find the equation, in slope-intercept form, of the line through each of the following pairs of points.

    (2,0) (3,4)

    1.Find the slope

    0-4= -4 reduced to 4
    2-3= -1
    M=4
    2. Use point slope form
    y-y1=M(x-x1)

    y-0=4(x-2)
    (simplfy)
    y-0=4x-8
    +0 +0 (Add 0 to both sides)

    Answer y=4x-8

    By Blogger Mkaladi, At September 20, 2007 at 6:48 PM  

  • Cera Per-4

    #24
    find slope-intp. for following points.
    (2,0)(3,4)

    First you need to find the slope of the two points.

    Y2-Y2
    ------
    X2-X1

    4-0 4
    ---- = --
    3-2 1

    Next find the Y intercept which you have to plug a point and we will use (2,0). PLug it into Y=mx+b and do the math.

    0=4(2)+b
    -8=b

    Now you use the info. we just found and plug in the numbers into the the slope-intp formula.

    y=mx+b

    Y=4x-8

    By Blogger Cera, At September 20, 2007 at 6:58 PM  

  • kayla period 4

    #5
    graph:
    y=5/6x-3

    to graph this problem, you will first need to find the y-intercept (b) and the slope (m).

    since they are already given to you, you can plug them into the slope intercept form y=mx+b.

    y=5/6x-3

    you can then make a term and value chart...

    x|y
    ---
    0|-3
    1|-13/6
    2|-4/3
    3|-1/2

    * i got the values above from:

    y=(5/6)(0)-3
    y=0-3
    y=-3

    y=(5/6)(1)-3
    y=5/6-3
    y=-13/6

    y=(5/6)(2)-3
    y=5/3-3
    y=-4/3

    y=(5/6)(3)-3
    y=5/2-3
    y=-1/2

    after you get your values, graph them and connect by using a stright edge.

    * it helps to check your graph by picking a point on the line and plugging it into the point-slope form y-y1=m(x-x1)

    By Blogger kayla's a veg head., At September 20, 2007 at 7:07 PM  

  • This comment has been removed by the author.

    By Blogger sam.axelrod, At September 20, 2007 at 7:15 PM  

  • 4.5 #20
    determine whether or not the points (5,10), (15, -10), and (0,20) are collinear.

    you can plot the points and see if they lie on the same line.

    or you can find the slope between 2 points by using the formula -
    Y1-Y2 over X1-X2

    between (5,10) and (15,-10)
    10-(-10) = 20 = 2
    5-15 =-10 = -1
    m = -2

    between (15,-10) and (0,20)
    -10 -20 = -30 = -2
    15 -0 = 15 = 1
    m = -2

    both the slopes are the same so these ponits are collinear

    samantha axelrod pd.4

    By Blogger sam.axelrod, At September 20, 2007 at 7:17 PM  

  • Danielle per.4
    #20
    Determine whether or not each set of points are collinear

    (5,10) (15,-10) (0,20)

    do the rise/run equation
    (y2-y1)/(x2-x1)

    if you do rise/run then you should get

    -10-10=-20
    15-5=10
    so -20/10=-2/1 reduced = -2

    20-(-10)=30
    0-15=-15
    so 30/-15=2/-1 reduced = -2


    since they have the same slope of
    -2 they are collinear.

    By Blogger Unknown, At September 20, 2007 at 7:21 PM  

  • 4.5 #26

    Find the equation, in slope-intercept form, of the line through each of the following pairs of points (1,2) (3,-4)

    Juan R Per.4

    = y=3x+5

    plug in the points into y1-y2 over x1-x2 to find m (slope)

    it would be 2-(-4)=6 and 1-3= -2
    which when simplified = -3 so m=-3

    plug that into point slope form =
    y-2=-3(x-1)

    to put it in slope intercept form just solve for y = y-2=-3x+1
    +2 +2
    = y=-3x+5

    By Blogger JuanR, At September 20, 2007 at 7:23 PM  

  • rodriguez per.3
    answer: y=x+1
    workings:
    2-4/1-3=1
    i graphed the points and made a line. the point that intercept the y axis is 1 so 1 is the y intercept.

    By Blogger rodriguez, At September 20, 2007 at 8:19 PM  

  • Amanda period 4

    #6

    you would graph that by realizing when y=4, 4 is the y-intercept and also, the slope. So a point would be on(0,4), and there would be anouther one on(1,8), because you go up 4 and over 1.

    By Blogger 4thp~Mandy_Cat, At September 20, 2007 at 8:37 PM  

  • Romina: Period 4

    #25
    Find the equation, in slope-intercept form, of the line through each of the following pairs or points.

    (1,2)(3,-4)

    1. Calculate slope..

    2-(-4)/1-3=
    6/-2 [simplyfy]
    -3 <<--[slope]

    2.Choice a piont to plug in for x,y and substitute m.

    y-2=-3(x-1)
    y-2=3x+3 [solve for y, and then combine like-terms]



    Answer-->> y=3x+5

    :]♥

    By Blogger romi.4, At September 20, 2007 at 9:00 PM  

  • This comment has been removed by the author.

    By Blogger romi.4, At September 20, 2007 at 9:01 PM  

  • spencer period 4.
    #11
    okay so your two points are (-4,5) and (4,5)
    Y1-Y2 over X1-X2 is what i used.
    so Y1=5
    Y2=5
    X1=-4
    X2=4
    so 5-5 over -4-4 equals 0 over -8
    and then you do the opposite recipical and you so you have M=-8


    then you find the midpoint and so i added 5+5 which equals 10 divided by 2 is 5 and thats your Y.

    and then 4+(-4) or 4-4 equals 0 divided by 2 is 0.
    so your midpoint is (0,5)

    then you use Y-Y1=M(X-X1)

    so you have Y-5=-8(X-0)

    and then you times -8 and X

    and you times -8 and 0

    so then you have Y-5=-8X

    then you add 5 to each side and you get your answer:

    Y=-3X

    By Blogger Molly.S., At September 20, 2007 at 9:29 PM  

Post a Comment

Subscribe to Post Comments [Atom]



<< Home