In the known triangle ABC, a (2, - 1), B (3,2), C (- 3, - 1), the height of the edge of BC is ad

In the known triangle ABC, a (2, - 1), B (3,2), C (- 3, - 1), the height of the edge of BC is ad

AD=(-1,2)
D (1,1)
This topic is just talking
Add a few auxiliary lines. Take your time
Let B be the vertical line of AC and foot E; let d be the vertical line of AC and foot f; let d be the vertical line and foot G
Then, according to the parallel proportion in some triangles and the similarity of right angles in ACD, it can be calculated slowly
Golden Archean
X = 0 is f (x) = (1 + x) ^ 1 / X what is the type of discontinuity!
According to the important limit, X tends to 0, and the positive and negative limits are both E, so they are removable discontinuities
Is it possible to go to the breakpoint? Ask: can you tell me more about it
Let {an} be a sequence of positive integers, the sum of the first n terms is Sn, and the mean of the equal difference between an and 2 is equal to the mean of the equal ratio between Sn and 2,
This problem needs a parody,
Because the mean difference between an and 2 is equal to the mean difference between Sn and 2, the square of (an + 2) / 2 = 2Sn, that is, (an + 2) ^ 2 = 8Sn, so (an-1 + 2) ^ 2 = 8sn-1, an ^ 2 + 4an-an-1 ^ 2-4an-1 ^ 2 = 8An, that is, an ^ 2-4an-an-1 ^ 2-4an-1 ^ 2 = 0, so (an ^ 2-an-1 ^ 2) - 4 (an + an-1) = 0, so (an + an-1) (an ^ 2-an-1 ^ 2-4) = 0, and because {an} is a sequence of positive integers, So (an + an-1) is not equal to 0, so (an ^ 2-an-1 ^ 2-4) = 0An is the arithmetic sequence with tolerance of 4, and A1 is equal to Sn and 2 by using the arithmetic mean of an and 2
So the general term of {an} can be found,
There is a column of numbers A1, A2, A3 Where: A1 = 6 × 2 + 1 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; A2 = 6 × 3 + 2A3 = 6 × 4 + 3 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; A4 = 6 × 5 + 4 Then the nth number an=______ (expressed in an algebraic expression containing n)
a1=6×2+1=6×（1+1）+1，a2=6×3+2=6×（2+1）+2，a3=6×4+3=6×（3+1）+3，a4=6×5+4=6×（4+1）+4，… Then the nth number an = 6 (n + 1) + n = 7n + 6
It is known that the three vertices of triangle ABC are a (2, - 1), B (3,2), C (- 3, - 1). If the height of BC side is ad, the coordinates of point D and AD are obtained
The slope of BC is (2 + 1) / (3 + 3) = 1 / 2
So the slope of ad is - 2
Over a
Ad is y + 1 = - 2 (X-2)
y=2x-5
BC over b
Is Y-2 = 1 / 2 * (x-3)
x-2y+1=0
Substituting y = 2x-5
x-4x+10+1=0
x=11/3,y=7/3
So D (11 / 3,7 / 3)
D(1,1)
Ad linear equation y = - 2x + 3
If ad is high, ad must be perpendicular to BC, which is the key
Let D (x, y), then
Ad vector = (X-2, y + 1)
Vector of BD = (x-3, Y-2)
The vector of CB = (6,3)
AD⊥BC,BD‖BC
6(x-2)+3(y+1)=0,3(x-3)=6(y-2)
x=1,y=1
D (1,1) Ad vector = (- 1,2)
Find the discontinuities of the following functions and state the type F (x) = {x + 1, x = 1
When x = 1, f (x) = 1
lim(x->1-)f(x)=1+1=2
So f (x) is discontinuous at x = 1
The left limit of ∵ f (x) at x = 1 is 1, the right limit is 2, and ∵ x = 1 is the jumping point of F (x).
The first kind of discontinuity is the continuous limit between two points
Let {an} be a sequence of positive numbers, the sum of the first n terms is Sn, and for all positive integers n, the mean of the equal difference between an and 2 is equal to the mean of the equal ratio between Sn and 2, find the general term formula of the sequence {an}
∵ the mean of the equal difference between an and 2 is equal to the mean of the equal ratio between Sn and 2, ∵ 12 (an + 2) = 2Sn, that is, Sn = 18 (an + 2) 2 (2 points) when n = 1, S1 = 18 (a1 + 2) 2 {A1 = 2; & nbsp (3) when n ≥ 2, an = Sn − Sn − 1 = 18 [(an + 2) 2 − (an − 1 + 2) 2], that is, (an + an-1
There is a column of numbers A1, A2, A3 If A1 = 2, then a2011 is ()
A. 2011B. 2C. -1D. 12
∵a1=2，∴a2=1-12=12，a3=1-2=-1，a4=1-（-1）=2，a5=1-12=12，… And so on, every three numbers for a group of circulation, 2011 △ 3 = 670 So the answer is: 2
It is known that the vertices of △ ABC are a (2,1), B (3,2), C (- 3, - 1), and the height of BC side is ad. the coordinates of Ad vector and point D are obtained
Let D (x, y)
∵ vector BC = oc-ob = (- 3-3, - 1-2) = (- 6, - 3), ad = (X-2, Y-1)
For BC ⊥ ad, as long as BC · ad = 0:2x + y = 5
Because D is on BC: y = 0.5x + 0.5, x = 1.8, y = 1.4
Let f (x) = LiMn →∞ (n − 1) xnx2 + 1, then the discontinuity of F (x) is X=______ .
It is obvious that when x = 0, f (x) = 0; when x ≠ 0, f (x) = LiMn →∞ (n − 1) xnx2 + 1 = xlimn →∞ 1 − 1nx2 + 1n = x · 1x2 = LX  f (x) = 0, x = 01x, X ≠ 0  limx → 0f (x) = infinity, so x = 0 is the discontinuity of F (x)