ArrayArray Get it done six.step 3 Medians and you will Altitudes of Triangles | Saludvidable

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

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

Concern 1. Language Identity the newest five variety of points off concurrency. And this contours intersect to create each of the points? Answer:

## Get the duration of the section

Question 2PLETE The fresh Phrase The size of a section out of a beneficial vertex into the centroid is actually ______________ the duration of the brand new median out-of you to definitely vertex.

Answer: The duration of a section off a great vertex to the centroid is one-3rd of one’s period of new average of that 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

Within the Teaching eleven-14. point G ‘s the centroid out of ?ABC. BG = six, AF = twelve, and you can AE = 15.

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

In the Knowledge 19-22. give whether or not the orthocenter is to the, with the, otherwise beyond your triangle. Upcoming discover coordinates of your 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 – wildbuddies 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.

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