Get it done six.step 3 Medians and you will Altitudes out of Triangles

Question step one. Code Identity the latest four variety of circumstances away from concurrency. And this lines intersect to make each one of the facts? Answer:

Question 2PLETE The new Phrase The length of a segment regarding a vertex toward centroid try ______________ along the brand new median out of one to vertex.

Answer: The duration of a segment away from a beneficial vertex to your centroid is one-third of your own length of new average regarding 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

Into the Exercises eleven-14. point G ‘s the centroid off ?ABC. BG = 6, AF = twelve, and you will AE = fifteen. Discover the amount of the fresh new sector.

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

In Knowledge 19-twenty-two. give perhaps the orthocenter is into the, for the, or away from triangle. Then select the coordinates of your own orthocenter.

Explanation: hater uygulaması nedir 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.