6.5 Secondary Proof and Inequalities in one Triangle

6.step three Medians and you can Altitudes away from Triangles

Share with if the orthocenter of the triangle toward offered vertices are into the, toward, otherwise outside the triangle. Upcoming discover 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 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 brand new Triangle Midsegment Theorem

Answer: The new midsegment of Ab = (-six, 6) New midsegment from BC = (-step 3, 4) The newest midsegment out of Air conditioning = (-step three, 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 fresh midsegment of De = (0, 3) The new midsegment off EF = (dos, 0) The fresh midsegment out-of DF = (-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 https://datingranking.net/tr/caribbean-cupid-inceleme/ < 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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