Do it 6.step three Medians and Altitudes out of Triangles
Question step one. Language Identity the five brand of circumstances of concurrency. Which traces intersect in order to create all the things? Answer:
Matter 2PLETE The brand new Sentence Along a section of good vertex into the centroid was ______________ the size of this new median out of one vertex.
Answer: The duration of a segment of an excellent vertex to the centroid is certainly one-third of one’s length of brand new average out of you to definitely vertex.
Explanation: The centroid of the trinagle = (\(\frac < 1> < 3>\), \(\frac < 4> < 3>\)) = (\(\frac < 10> < 2>\), 3)
Explanation: PN = \(\frac < 2> < 3>\)QN PN = \(\frac < 2> < 3>\)(21) PN = 14 QP = \(\frac < 1> < 3>\)QN = \(\frac < 1> < 3>\)(21) = 7
Explanation: PN = \(\frac < 2> < 3>\)QN PN = \(\frac < 2> < 3>\)(42) PN = 28 QP = \(\frac < 1> < 3>\)QN = \(\frac < 1> < 3>\)(42) = 14
Explanation: DE = \(\frac < 1> < 3>\)CE 11 = \(\frac < 1> < 3>\) CE CE = 33 CD = \(\frac < 2> < 3>\) CE CD = \(\frac < 2> < 3>\)(33) CD = 22
Explanation: DE = \(\frac < 1> < 3>\)CE 15 = \(\frac < 1> < 3>\) CE CE = 45 CD = \(\frac < 2> < 3>\) CE CD = \(\frac < 2> < 3>\)(45) CD = 30
During the Teaching 11-fourteen. section Grams is the centroid regarding ?ABC. BG = 6, AF = several, and AE = 15. Get the period of the portion.
Explanation: The centroid of the trinagle = (\(\frac < 1> < 3>\), \(\frac < 5> < 3>\)) = (\(\frac < -7> < 3>\), 5)
Explanation: DE = \(\frac < 1> < 3>\)CE 11 = \(\frac < 1> < 3>\) CE CE = 33 CD = \(\frac < 2> < 3>\) CE CD = \(\frac < 2> < 3>\)(33) CD = 22
Inside Training 19-twenty-two. tell if the orthocenter is to the, into, or outside of the triangle. Next find the coordinates of your own orthocenter.
Explanation: The slope of YZ = \(\frac < 6> < -3>\) = \(\frac < -1> < 2>\) The slope of the perpendicular line is 2 The equation of perpendicular line is (y – 2) = 2(x + 3) y – 2 = 2x + 6 2x – y + 8 = 0 The slope of XZ = \(\frac < 6> < -3>\) = 0 The equation of perpendicular line is (y – 2) = 0 y = 2 Substitute y = 2 in 2x – y + 8 = 0 2x – 2 + 8 = 0 2x + 6 = 0 x = -3 the orthocenter is (-3, 2) The orthocenter lies on the vertex of the triangle.
Explanation: The slope of UV = \(\frac < 4> < 0>\) = \(\frac < -3> < 2>\) The slope of the perpendicular line is \(\frac < 2> < 3>\) The equation of the perpendicular line is (y – 1) = \(\frac < 2> < 3>\)(x + 2) 3(y – 1) = 2(x + 2) 3y – 3 = 2x + 2 2x – 3y + 5 = 0 — (i) The slope of TV = \(\frac < 4> < 0>\) = \(\frac < 3> < 2>\) The slope of the perpendicular line is \(\frac < -2> < 3>\) The equation of the perpendicular line is (y – 1) = \(\frac < -2> < 3>\)(x – 2) 3(y – 1) = -2(x – 2) 3y – 3 = -2x + 4 2x + 3y – 7 = 0 -(ii) Add two equations 2x – 3y + 5 + 2x + 3y – 7 = 0 4x – 2 = 0 x = 0.5 2x – 1.5 + 5 = 0 x = -1.75 So, the orthocenter is (0, 2.33) The orthocenter lies inside the triangle ABC.