anonymous
  • anonymous
The line containing the longer diagonal of a quadrilateral whose vertices are A (2, 2), B(-2, -2), C(1, -1), and D(6, 4).
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
calculate AC and BD using distance formula and the longer one will be the line containing the longer diagonal of the quadrilateral
anonymous
  • anonymous
so x-y=9?
anonymous
  • anonymous
or 2x+1y=9

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anonymous
  • anonymous
AC = sqrt[(x2 - x1)^2 + (y2 - y1)^2]
anonymous
  • anonymous
AC = sq.rt[(1 - 2)^2 + (-1 -2)^2] ac = sqrt[(-1)^2 + (-3)^2] AC = sqrt[1 + 9] ac = sqrt(10)
anonymous
  • anonymous
similarly calculate BD
anonymous
  • anonymous
B(-2, -2), D(6, 4). BD = sq.rt{[6 - (-2)]^2 + [4 -(-2)]^2} BD = sqrt[(6 + 2)^2 + (4 + 2)^2] BD = sqrt[(8)^2 + (6)^2] BD = sqrt(64 + 36) BD = sqrt(100) BD = 10 so BD is the longer one
anonymous
  • anonymous
can you help me with another equation
anonymous
  • anonymous
sure, go ahead..
anonymous
  • anonymous
Indicate the equation of the given line in standard form. The line containing the median of the trapezoid whose vertices are R(-1, 5) , S(l, 8), T(7, -2), and U(2, 0). This is the question... and for the answer i got was: 13+x-y=48? can you tell me where i went wrong
anonymous
  • anonymous
We can first find the midpoints of the legs RU and ST using the Midpoint formula of a line segment. Then we find the distance between them by using the formula for Distance between Two Points.
anonymous
  • anonymous
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anonymous
  • anonymous
R(-1, 5) , U(2, 0). midpoint of RU = X and its coordinates are -1 + 2 5 + 0 x = -------- and y = -------- so X = (1/2 , 5/2) 2 2 S(l, 8), T(7, -2), midpoint of ST = Y and its coordinates are 1 + 7 8 + (-2) x = ------- and y = --------- so Y = (4, 3) 2 2 so we have to find the line passing through the points x (1/2, 5/2) and y(4, 3)
anonymous
  • anonymous
so we have to find the line passing through the points X= (1/2, 5/2) and Y=(4, 3) First we find the slope of the line y2 - y1 3 - 5/2 1/2 m = --------- = --------- = -------- = 1/7 x2 - x1 4 - 1/2 7/2 Now we use the Point-slope form and using one of the points, say Y, we find the equation of the line y - y1 = m(x - x1) y - 3 = 1/7(x - 4) y - 3 = x/7 - 4/7 Now expressing it in standard form -3 + 4/7 = x/7 - y x/7 - y = -17/7 so eqn can be 1 17 --- x - y = - ----- 7 7 or x - 7y = -17

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