Given the triangle ABC, the coordinates of three vertices are a (- 1,0), B (1,2), C (0, x), and the vector AB is perpendicular to the vector BC, then the value of C is

Given the triangle ABC, the coordinates of three vertices are a (- 1,0), B (1,2), C (0, x), and the vector AB is perpendicular to the vector BC, then the value of C is

According to the vector vertical formula, X1 * x2 + Y1 * y2 = 0, and the vector AB = (2,2) BC (- 1, X-2), i.e
-2 + 2 (X-2) = 0 leads to x = 3, so the coordinate of C = (0,3)
Given that the coordinates of the three vertices of the triangle ABC are a (- 2,3) B (1,2) C (5,4), find the vector BA.BC Detailed process of coordinate () of
Classmate. Let me first explain that, for example, the vector AB = (x, y) is not called the coordinates of a vector, but the vector (x, y) is the expression of a vector, and only points have coordinates
Vector Ba = coordinates of point B - coordinates of point a = (1 - (- 2), 2-3) = (3, - 1)
Similarly, the vector BC = (4,2)
Mathematics! One side of △ ABC is fixed, and vertex C moves on a fixed line L parallel to ab. let the perpendicular center of △ ABC be in the triangle, and find the trajectory equation of the perpendicular center
Urgent! Do it with parametric equation!
Take the line where AB is as the x-axis and the midpoint o of AB as the origin as the coordinate system xoy
Let a (- 1,0), B (1,0). L: y = L, C (C, l)
Let CF ⊥ AB, f; ad ⊥ BC, D; be ⊥ AC, e
The center of perpendicularity g (x, y) is the intersection of CF and AD
CF：x=c
The center of perpendicularity is in △ ABC, so - 1 ≤ x ≤ 1
Slope of BC = L / (C-1)
So the slope of ad = (1-C) / L
AD:y=(1-c)/[l*(x+1)]
The intersection trajectory of AD and CF is: L * y = (1-x) / (1 + x)
In the process, l is the lowercase of L, which may not be clear
The known sequence {an} is an arithmetic sequence with non-zero tolerance, A1 = 1, and A2, A4, a8 are equal proportion sequence
② If BN = 1 / [an * a (n + 1)], find the first n terms of {BN} and Sn
Let an = a1 + (n-1) d
That is, an = 1 + (n-1) d
A2, A4 and A8 are equal ratio series
(1+3d)^2=(1+d)(1+7d)
1+6d+9d^2=1+8d+7d^2
2d^2=2d
d≠0 d=1
an=1+(n-1)=n
bn=1/[n(n+1)]
Sn=1/(1*2)+1/(2*3)+1/(3*4)+.+ 1/[n(n+1)]
=1/1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)
=1-1/(n+1)=n/(n+1)
Given the vertex a (- 3,0), B (- 1, - 4) of triangle ABC, vertex C moves on the line 2x-y-5 = 0, find the trajectory equation of the center of gravity P of triangle ABC
Given the vertex a (- 3,0), B (- 1, - 4) of △ ABC and the vertex C moving on the line 2x-y-5 = 0, the trajectory equation of the center of gravity P of △ ABC is obtained
Kaka-
The time limit is only 2 hours
If it's OK, there will be points
Ca, can you explain --
Kaka^^
Let p be (x, y) and C be (m, n)
-3-1+m=3x
0-4+n=3y
So there are: M = 3x + 4, n = 3Y + 4
And C is on the line 2x-y-5 = 0, so there is:
2(3x+4)-(3y+4)-5=0
6x+8-3y-4-5=0
The trajectory equation is: 6x-3y-1 = 0
In the equal difference sequence {an} and equal ratio sequence {BN} with tolerance D ≠ 0, given A1 = 1, A1 = B1, a8 = B3, find the first n terms and Sn of {an + BN}
Sn = (6N power - 1) / 5 + (5N square - 3n) / 2
The vertex a of triangle ABC is fixed, the opposite side BC of point a is 2a, the length of the height on the side BC is B, and the side BC moves along a certain straight line to find the locus of the outer center of the triangle
Do the rectangular coordinate system x0y
Take point a (0, b), BC edge moves along X axis
Let the coordinates of outer center (x, y), B point (x1,0), C point (x1-2a,0)
It can be seen from the outer center that the distance from the center of the circle to the three points is equal
Get the equations: (the following are square)
X-square + (y-b) square = y-square + (x-x1) square = y-square + (x-x1 + 2a) square
The solution of the system of square equations is: B + a - by
Take a as the origin, the height on the side of BC as the x-axis, and the straight line where BC is located as x = B, establish a rectangular coordinate system. Let the outer center of △ ABC be p (x, y), and p be on the vertical bisector of BC, then the coordinates of B and C are (x + A, 0), (x-a, 0) respectively
And: ∥ AP ∥ = ∥ BP ∥
√(x^2+y^2)=√[a^2+(y-b)^2]
That is, the trajectory of point P is: x ^ 2 = - 2by + A ^ 2-B ^ 2
It is known that the tolerance of the arithmetic sequence {an} d = 1 + and the sum of the first n terms is SN. If 1, A1 and A3 are in an equal proportion sequence, A1 can be obtained
If S5 > a1a9, find the range of A1
a(n) = a + (n-1),
a(1)=a,
[a(1)]^2 = a^2 = 1*a(3) = a+2,
0 = a^2 - a - 2 = (a-2)(a+1),
A (1) = a = 2 or a (1) = a = - 1
s(n) = na + n(n-1)/2,
a(1)a(9) = a(a+8) a^2 + 3a - 10 = (a+5)(a-2),
-5 < a = a(1) < 2.
Given ABC vertices a (3,3), B (3, - 3), vertex C moves on the circle x ^ 2 + y ^ 2 = 9, find the trajectory equation of the center of gravity g of triangle ABC
Let the coordinates of point C be C (x, y). Let G be g (x, y). Let d be d (3,0). Then: x = 3 - (3-x) / 3 = 2 + X / 3 -------- 1y = Y / 3 -------- 2
Let (an) be an arithmetic sequence, A1 = 1, Sn be the sum of the first n terms, and (BN) be an arithmetic sequence whose absolute value of common ratio q is less than 1
It is known that the sequence {an} is an arithmetic sequence, A1 = 1, Sn is the sum of its first n terms: {BN} is an arithmetic sequence, the absolute value of its common ratio q is less than 1, TN is the sum of its first n terms, if A4 = B2, S6 = 2t2-1, limtn = 8, find the general formula of {an} and {BN}
From A4 = B2: 1 + 3D = b1q
From S6 = 2t2-1, 6 + 15d = 2B1 (1 + Q) - 1
From limtn = 8, B1 / (1-Q) = 8
Solve d, B1 and Q