Answer:
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
