1.)indicate the equation of the given line in standard form. show all of your work for full credit.

the line containing the median of the trapezoid whose vertices are r(-1, 5) , s(1, 8), t(7, -2), and u(2, 0).

2.)indicate the equation of the given line in standard form. show all of your work for full credit.

the line containing the altitude to the hypotenuse of a right triangle whose vertices are p(-1, 1), q(3, 5), and r(5, -5).

3.) indicate the equation of the given line in standard form. show all of your work for full credit.

the line containing the diagonal, bd, of a square whose vertices are a(-3, 3), b(3, 3), c(3, -3), and d(-3, -3). find two equations, one for each diagonal.

1) The equation of the line containing the median of

   trapezium is 10x + 6y = 39 ⇒ (standard form)

2) The equation of the line containing the altitude to the

    hypotenuse of a right angle triangle is x - 5y = -6

3) The equations of the lines containing the diagonals

    BD is y = x  and AC is y = -x

Step-by-step explanation:

1) ∵ RSTU is a trapezium where

      R (-1 , 5) , S (1 , 8) , T (7 , -2) , U (2 , 0)

∴ It has 2 parallel bases

* Lets use the Rule of slope to find the parallel bases

  because the parallel lines equal in their slopes

∵ Slope of RS = (8 - 5)/(1 - -1) = 3/2

∵ Slope of ST = (-2 - 8)/(7 - 1) = -10/6 = -5/3

∵ Slope of TU = (0 - -2)/(2 - 7) = 2/-5 = -2/5

∵ Slope of Ru = (0 - 5)/(2 - -1) = -5/3

∵ Slope of ST = Slope of RU

∴ ST // RU

∵ The median of trapezium is parallel to ST and RU

∴ It has the same slope

∴ The slope of the median base is -5/3

∵ The median base of the trapezium is passing through

   the mid-points of nonparallel bases

∵ The mid-point of RS = [(-1 + 1)/2 , (5 + 8)/2] = (0 , 13/2)

∵ The mid-point of TU = [(7 + 2)/2 , (-2 + 0)/2] = (9/2 , -1)

* Now we can find the equation of the median base using

  its slope and  point lies on it

∵ (y  - y1) = m(x - x1) ⇒ m = -5/3 , (x1 , y1) = (0 , 13/2)

∴ (y - 13/2) = (-5/3)(x - 0)

∴ y - 13/2 = -5/3 x ⇒ multiply both sides by 6

∴ 6y - 39 = -10x

∴ 10x + 6y = 39 ⇒ standard form

2) At first we must find the slopes of PQ , QR , PR to

   find the perpendicular sides and the hypotenuse

∵ P (-1 , 1) , Q (3 , 5) , R (5 , -5)

∵ The slope of PQ = (5 - 1)/(3 - -1) = 4/4 = 1

∵ The slope of QR = (-5 - 5)/(5 - 3) = -10/2 = -5

∵ The slope of PR = (-5 - 1)/(5 - -1) = -6/6 = -1

∵ The product of the ⊥ slopes is -1

∵ Slope PQ × Slope PR = 1 × -1 = -1

∴ PQ ⊥ PR ⇒ ∠P is a right angle

∴ QR is the hypotenuse with slope -5

∵ The altitude from the right angle to the hypotenuse

   has slope = -5 × m = -1 ⇒ m = -1/-5 = 1/5

∵ It passing through point P

∴ Its equation is:

  (y - 1) = 1/5(x - -1) ⇒ y - 1 = 1/5 x + 1/5

∴ y = 1/5 x + 1/5 + 1

∴ y = 1/5 x + 6/5 ⇒ multiply both sides by 5

∴ 5y = x + 6

∴ x - 5y = -6 ⇒ standard form

3) ∵ ABCD is a square with vertices

   A (-3 , 3) , B (3 , 3) , C (3 , -3) , D (-3 , -3)

∵ Its diagonal ⊥ to each other

∴ AC ⊥ BD

∴ The product of their slopes = -1

∴ The slope of BD = -3 - 3/-3 - 3 = -6/-6 = 1

∴ The slope of AC = -1

∵ The equation of BD is (y - 3) = 1( x - 3)

∴ y - 3 = x - 3

∴ y = x ⇒ standard form

∵ The equation of AC is (y - 3) = -1( x - -3)

∴ y - 3 = -x - 3

∴ y = -x ⇒ standard form

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