Explanation: The centroid of the trinagle = (\(\frac < 1> < 3>\), \(\frac < 4> < 3>\)) = (\(\frac < 10> < 2>\), 3)
Concern step 1. Vocabulary Identity the newest five brand of circumstances away from concurrency. Which traces intersect to create all the situations? Answer:
Select the length of the newest section
Matter 2PLETE The new Phrase Along a section off good vertex on the centroid was ______________ along this new average of you to vertex.
Answer: The duration of a segment away from a vertex to your centroid is the one-3rd of one’s length of brand new median regarding one to vertex.
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
In the Knowledge eleven-14. part Grams is the centroid regarding ?ABC. BG = six, AF = 12, and you can AE = 15.
Explanation: The centroid of the trinagle = (\(\frac < 1> < 3>\), \(\frac < 5> < 3>\)) = (\(\frac < -7> < 3>\), 5)
Into the Training 19-22. tell whether or not the orthocenter is to the, on, otherwise beyond your triangle. Following get the coordinates of the 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 sitio de citas para reclusos 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.