A median of a triangle is a line segment joining a vertex of a triangle to the midpoint of the opposite side. The three medians of a triangle are drawn below. [asy] pair A, B,C, X,Y, Z; A = (0,0); B = (1,0); C = (0.3,0.8); X = (B+C)/2; Y = (A+C)/2; Z = (A+B)/2; draw(A–X, red); draw(B–Y, red); draw(C–Z, red); draw(A–B–C–A); [/asy] Note that the three medians appear to intersect at the same point! Let’s try this out with a particular triangle. Consider the triangle $ABC$ with $A = (3,6)$, $B = (-5,2)$, and $C = (7,-8)$. (a) Let $D,$ $E,$ $F$ be the midpoints of $overline{BC},$ $overline{AC},$ $overline{AB},$ respectively. Find the equations of medians $overline{AD},$ $overline{BE},$ and $overline{CF}.$ (b) Show that the three medians in part (a) all pass through the same point.

Posted on By mobete 4 Comments on A median of a triangle is a line segment joining a vertex of a triangle to the midpoint of the opposite side. The three medians of a triangle are drawn below. [asy] pair A, B,C, X,Y, Z; A = (0,0); B = (1,0); C = (0.3,0.8); X = (B+C)/2; Y = (A+C)/2; Z = (A+B)/2; draw(A–X, red); draw(B–Y, red); draw(C–Z, red); draw(A–B–C–A); [/asy] Note that the three medians appear to intersect at the same point! Let’s try this out with a particular triangle. Consider the triangle $ABC$ with $A = (3,6)$, $B = (-5,2)$, and $C = (7,-8)$. (a) Let $D,$ $E,$ $F$ be the midpoints of $overline{BC},$ $overline{AC},$ $overline{AB},$ respectively. Find the equations of medians $overline{AD},$ $overline{BE},$ and $overline{CF}.$ (b) Show that the three medians in part (a) all pass through the same point.

A median of a triangle is a line segment joining a vertex of a triangle to the midpoint of the opposite side. The three medians of a triangle are drawn below. [asy] pair A, B,C, X,Y, Z; A = (0,0); B = (1,0); C = (0.3,0.8); X = (B+C)/2; Y = (A+C)/2; Z = (A+B)/2; draw(A--X, red); draw(B--Y, red); draw(C--Z, red); draw(A--B--C--A); [/asy] Note that the three medians appear to intersect at the same point! Let's try this out with a particular triangle. Consider the triangle $ABC$ with $A = (3,6)$, $B = (-5,2)$, and $C = (7,-8)$. (a) Let $D,$ $E,$ $F$ be the midpoints of $\overline{BC},$ $\overline{AC},$ $\overline{AB},$ respectively. Find the equations of medians $\overline{AD},$ $\overline{BE},$ and $\overline{CF}.$ (b) Show that the three medians in part (a) all pass through the same point.

Mathematics

Comments (4) on “A median of a triangle is a line segment joining a vertex of a triangle to the midpoint of the opposite side. The three medians of a triangle are drawn below. [asy] pair A, B,C, X,Y, Z; A = (0,0); B = (1,0); C = (0.3,0.8); X = (B+C)/2; Y = (A+C)/2; Z = (A+B)/2; draw(A–X, red); draw(B–Y, red); draw(C–Z, red); draw(A–B–C–A); [/asy] Note that the three medians appear to intersect at the same point! Let’s try this out with a particular triangle. Consider the triangle $ABC$ with $A = (3,6)$, $B = (-5,2)$, and $C = (7,-8)$. (a) Let $D,$ $E,$ $F$ be the midpoints of $overline{BC},$ $overline{AC},$ $overline{AB},$ respectively. Find the equations of medians $overline{AD},$ $overline{BE},$ and $overline{CF}.$ (b) Show that the three medians in part (a) all pass through the same point.”

  1. (a) The equation of AD is y = 4.5·x -7.5

    The equation of BE is y = -0.3·x + 0.5

    The equation of CF is y = -1.5·x + 2.5.

    (b) The three equations representing the three lines AD, BE and CF, pass through the same point [tex](1\frac{2}{3} , 0)[/tex] (which can be written as (1.667, 0)).

    Step-by-step explanation:

    The given triangle coordinates are;

    A = (3, 6)

    B = (-5, 2)

    C = (7, -8)

    The midpoint of BC = D

    The midpoint of AC = E

    The midpoint of AB = F

    Therefore, we have;

    The coordinates of point D = (B + C)/2 = (((-5) + 7 )/2, (2 + (-8))/2)

    The coordinates of point D =  (1, -3)

    The coordinates of point E = (A + C)/2 = ((3 + 7 )/2, (6 + (-8))/2)

    The coordinates of point E =  (5, -1)

    The coordinates of point F = (A + B)/2 = ((3 + (-5) )/2, (6 + 2)/2)

    The coordinates of point F =  (-1, 4)

    The equation of AD in slope and intercept form is therefore;

    [tex]Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

    A = (3, 6), D =  (1, -3)

    [tex]m =\dfrac{(-3)-6}{1-3} = 4.5[/tex]

    Therefore, we have;

    y - 6 = 4.5×(x - 3)

    y - 6 = 4.5·x -13.5

    y = 4.5·x -13.5 + 6 = 4.5·x -7.5

    y = 4.5·x -7.5

    The equation of BE in slope and intercept form is found as follows;

    B = (-5, 2), E =  (5, -1)

    Slope = ((-1)-2)/(5 - (-5)) = -0.3

    The point slope equation is y - 2 = -0.3×(x - (-5))

    y - 2 = -0.3·x - 1.5

    y = -0.3·x - 1.5 + 2 = -0.3·x + 0.5

    y = -0.3·x + 0.5

    The equation of CF in slope and intercept form is found as follows;

    C = (7, -8), F =  (-1, 4)

    Slope = (4 - (-8))/((-1) - 7) = -1.5

    The point slope equation is y - (-8) = -1.5×(x - 7)

    y + 8 = -1.5·x + 10.5

    y = -1.5·x + 10.5 - 8

    y = -1.5·x + 2.5.

    Where CF and BE intersect, we have;

    -0.3·x + 0.5 = -1.5·x + 2.5.

    1.5·x -0.3·x = 2.5 - 0.5 = 2.0

    x = 2.0/1.2 = 5/3

    y = -1.5×5/3+ 2.5 = 0

    We check for line AD, where y = 4.5·x -7.5

    When x = 5/3, we get;

    y = 4.5×(5/3) -7.5  = 0

    Therefore, the three equations pass through the point (5/3, 0) which is the same point.

    Reply

  2. answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

    step-by-step explanation:

    the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

    Reply

Leave a Reply

Your email address will not be published.