Take action six.3 Medians and Altitudes regarding Triangles

Take action six.3 Medians and Altitudes regarding Triangles

Explanation: The centroid of the trinagle = (\(\frac < 1> < 3>\), \(\frac < 4> < 3>\)) = (\(\frac < 10> < 2>\), 3)

Matter step 1. Code Term brand new four sorts of points from concurrency. And therefore contours intersect in order to create each one of the points? Answer:

Find the period of the section

Matter 2PLETE The latest Sentence The length of a section out of good vertex for the centroid try ______________ the length of the newest median away from you to vertex.

Answer: The length of a section off a beneficial vertex to your centroid is but one-third of period of this new median regarding you 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

From inside the Teaching 11-14. point Grams ‘s the centroid out-of ?ABC. BG = 6, AF = twelve, and you may AE = 15.

Explanation: The centroid of the trinagle = (\(\frac < 1> < 3>\), \(\frac < 5> < 3>\)) = (\(\frac < -7> < 3>\), 5)

When you look at the Knowledge 19-22. tell whether or not the orthocenter is inside, into the, otherwise outside of the triangle. After that discover coordinates of one’s 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 randki xmeeting lies inside the triangle ABC.

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