6.3 Medians and you will Altitudes from Triangles
Share with perhaps the orthocenter of your own triangle into given vertices are inside, for the, otherwise outside the triangle. Up coming discover coordinates of the orthocenter.
Explanation: The slope of the line HJ = \(\frac < 1> < 3>\) = \(\frac < 5> < 2>\) The slope of the perpendicular line = \(\frac < -2> < 5>\) The perpendicular line is (y – 6) = \(\frac < -2> < 5>\)(x – 1) 5(y – 6) = -2(x – 1) 5y – 30 = -2x + 2 2x + 5y – 32 https://datingranking.net/pl/vanilla-umbrella-recenzja/ = 0 – (i) The slope of GJ = \(\frac < 1> < 3>\) = \(\frac < -5> < 2>\) The slope of the perpendicular line = \(\frac < 2> < 5>\) The equation of perpendicular line (y – 6) = \(\frac < 2> < 5>\)(x – 5) 5(y – 6) = 2(x – 5) 5y – 30 = 2x – 10 2x – 5y + 20 = 0 – (ii) Equate both equations 2x + 5y – 32 = 2x – 5y + 20 10y = 52 y = 5.2 Substitute y = 5.2 in (i) 2x + 5(5.2) – 32 = 0 2x + 26 – 32 = 0 2x = 6 x = 3 The orthocenter is (3, 5.2) The orthocenter lies inside the triangle.
Explanation: The slope of LM = \(\frac < 5> < 0>\) = \(\frac < 1> < 3>\) The slope of the perpendicular line = -3 The perpendicular line is (y – 5) = -3(x + 8) y – 5 = -3x – 24 3x + y + 19 = 0 — (ii) The slope of KL = \(\frac < 3> < -6>\) = -1 The slope of the perpendicular line = \(\frac < 1> < 2>\) The equation of perpendicular line (y – 5) = \(\frac < 1> < 2>\)(x – 0) 2y – 10 = x — (ii) Substitute (ii) in (i) 3(2y – 10) + y + 19 = 0 6y – 30 + y + 19 = 0 7y – 11 = 0 y = \(\frac < 11> < 7>\) x = -6 The othrocenter is (-6, -1) The orthocenter lies outside of the triangle
6.cuatro The fresh Triangle Midsegment Theorem
Answer: This new midsegment out-of Ab = (-6, 6) The newest midsegment off BC = (-step three, 4) The newest midsegment from Ac = (-step 3, 6)
Explanation: The midsegment of AB = (\(\frac < -6> < 2>\), \(\frac < 8> < 2>\)) = (-6, 6) The midsegment of BC = (\(\frac < -6> < 2>\), \(\frac < 4> < 2>\)) = (-3, 4) The midsegment of AC = (\(\frac < -6> < 2>\), \(\frac < 8> < 2>\)) = (-3, 6)
Answer: Brand new midsegment off De- = (0, 3) The newest midsegment regarding EF = (2, 0) The brand new midsegment regarding DF = (-step one, -2)
Explanation: The midsegment of DE = (\(\frac < -3> < 2>\), \(\frac < 1> < 2>\)) = (0, 3) The midsegment of EF = (\(\frac < 3> < 2>\), \(\frac < 5> < 2>\)) = (2, 0) The midsegment of DF = (\(\frac < -3> < 2>\), \(\frac < 1> < 2>\)) = (-1, -2)
Explanation: 4 + 8 > x 12 > x 4 + x > 8 x > 4 8 + x > 4 x > -4 4 < x < 12
Explanation: 6 + 9 > x 15 > x 6 + x > 9 x > 3 9 + x > 6 x > -3 3 < x < 15
Explanation: 11 + 18 > x 29 > x 11 + x > 18 x > 7 18 + x > 11 x > -7 7 < x < 29