six.step three Medians and you may Altitudes out-of Triangles
Share with perhaps the orthocenter of the triangle to your offered vertices is in to the, toward, otherwise beyond your triangle. Following get the coordinates of 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 = 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 https://datingranking.net/tr/amino-inceleme/ 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 new Triangle Midsegment Theorem
Answer: This new midsegment out-of Ab = (-6, 6) Brand new midsegment from BC = (-3, 4) This new midsegment out-of Air cooling = (-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: The newest midsegment off De = (0, 3) Brand new midsegment regarding EF = (2, 0) This new midsegment out-of 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
