Do it 6.step three Medians and you will Altitudes out-of Triangles

Concern step one. Language Title brand new five particular factors away from concurrency. And this lines intersect in order to create each one of the issues? Answer:

Matter 2PLETE The newest Sentence The size of a segment regarding an effective vertex into the centroid are ______________ the size of the new median from you to vertex.

Answer: The length of a segment regarding a vertex into the centroid is one-third of your own amount of this new median out-of 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 Knowledge eleven-fourteen. part G ‘s the centroid from ?ABC. BG = 6, AF = several, and AE = 15. Discover length of the new phase.

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 Practise 19-twenty-two. give whether or not the orthocenter is actually into the, toward, otherwise beyond your triangle. Then discover the 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 happn abonelik iptali, 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.