The three vertex coordinates a (2,1) B (0,3) C (- 1,2) ad of triangle ABC are the midline on BC. Find the vector ad coordinates

The three vertex coordinates a (2,1) B (0,3) C (- 1,2) ad of triangle ABC are the midline on BC. Find the vector ad coordinates

D=B+C/2=(-1/2,5/2)
Vector ad = D-A = (- 5 / 2,3 / 2)

In triangle ABC, the coordinates of vertex a are (2, - 1), the coordinates of B are (3,2), the coordinates of C are (- 3, - 1), and ad is the height on the edge of BC. Find the vector AD and point D

Let D coordinate be (x, y), then CB = (3,2) - (- 3, - 1) = (6,3) Da = (x, y) - (2, - 1) = (X-2, y + 1) and ad • CB = 0, that is, (X-2, Y-1) • (6,3) = 06x-12 + 3Y + 3 = 02x + Y-9 = 0... (1) let the linear equation of BC be y = KX + B, and substitute (3,2) (- 3, - 1) into 2Y = x + 1... (2

In △ ABC, if a (- 1,1), B (3,1), C (2,5), the bisector of the inner corner of angle a intersects the opposite edge of D, then the vector The coordinates of AD are equal to _

Let D (x, y), then AC=
(2+1)2+(5-1)2=5，AB=4，
According to the bisector theorem of triangle inner angle, BD = 4
5DC，
Namely:
BD=4
five
DC．
（x-3，y-1）=4
5（2-x，5-y），
∴
x-3=4
5(2-x)
y-1=4
5(5-y) ，
Solution
x=22
nine
y=25
nine ，
AD=(32
9，16
9)．
So the answer is: (32
9，16
9)．

It is known that the three vertices of triangle ABC are a (2, - 1), B (3,2), C (- 3, - 1), and the height on the edge of BC is ad. find the coordinates of point D and vector ad How do I do it in the vector method?

Let D (a, b) vector ad = (A-2, B + 1) vector CB = (6,3) vector CD = (a + 3, B + 1) vector DB = (3-A, 2-B)
Vector ad * vector CB = 6 (A-2) + 3 (B + 1) = 0 2A + B = 3
D on CB (a + 1) / (B + 3) = (2 + 1) / (3 + 3) 2A = B + 1
a=1 b=1
D(1,1)
Vector ad = (- 1,2)

In △ ABC, A.B.C is the length of the opposite side of angle A.B.C, cos B = 3 / 5, and vector ab × Vector BC = - 21 ① Find the area of △ ABC ② If a = 7, find angle C

a=BC,
c=AB,
∵ vector AB * vector BC = | a | * | C | * (- CoSb) = - 21
cosB=3/5
∴|a|*|c|=35
SINB = root [1 - (CoSb) ^ 2] = 4 / 5
S=1/2*|a|*|c|*sinB=1/2*35*4/5=14
∵a=7,∴c=5,
Cosine theorem: B ²= a ²+ c ²- 2a*c*cosB=32,∴b=4√2
Sine theorem C / sinc = B / SINB
sinC=√2/2,
And a > C, ∠ a > ∠ C, ∠ C = 45 °

In △ ABC, the opposite sides of ∠ a, ∠ B and ∠ C are a, B and C respectively, and meet Cos2 / a = 2 √ 5 / ab × Vector AC = 3 Find the area of (1) △ ABC (2). If B + C = 6, find the value of A And · meet Cos2 / a = 2 √ 5 / 5, ab × Vector AC = 3

Solution Cos2 / a = 2 √ 5 / 5
Then cosa = 2cos ² A-1=3/5
I.e. Sina = 4 / 5
By ab × Vector AC = 3
I.e. / AB / * / AC / * cosa = 3
I.e. / AB / * / AC / = BC = 3 / cosa = 5
That is, the area of △ ABC = 1 / 2 * / AB / * / AC / * Sina = 1 / 2 * 5 * 4 / 5 = 2
2 a ²= b ²+ c ²- 2bccosA
=（b+c） ²- 2bc-2bccosA
=（6） ²- 2*5-2*5*3/5
=20
I.e. a = 2 √ 5