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six.5 Secondary Facts and you will Inequalities in one Triangle

six.5 Secondary Facts and you will Inequalities in one Triangle

six.step 3 Medians and you will Altitudes regarding Triangles

Give if the orthocenter of triangle towards provided vertices is actually to the, on, or outside the triangle. After that discover the 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 = 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 – https://datingranking.net/de/bhm-dating-de 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

six.cuatro New Triangle Midsegment Theorem

Answer: The newest midsegment out-of Ab = (-six, 6) The brand new midsegment of BC = (-step 3, 4) The brand new midsegment out of Air-con = (-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 out of De = (0, 3) The new midsegment out of EF = (2, 0) The brand new midsegment regarding DF = (-step 1, -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

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