Home » Take action 6.step 3 Medians and you can Altitudes of Triangles

# Take action 6.step 3 Medians and you can Altitudes of Triangles

Take action 6.step 3 Medians and you can Altitudes of Triangles

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

Question 1. Words Label new five style of points out of concurrency. Which lines intersect to create all the facts? Answer:

## Get the period of the new segment

Concern 2PLETE The brand new Sentence The size of a section away from an effective vertex towards the centroid are ______________ the length of the newest median out of one vertex.

Answer: Along a segment away from an effective vertex to your centroid is the one-third of your amount of the latest average off one 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 Exercises eleven-14. point G ‘s the centroid off ?ABC. BG = 6, AF = a dozen, and you may AE = 15.

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

Within the Practise 19-22. tell if the orthocenter is in to the, for the, or outside the triangle. Following find 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 sesso incontri herpes 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.