The coordinates of the vertex B and C of the triangle ABC are (0,0) and (4,0) AB, and the length of the midline on the edge of (4,0) AB is 3

Let a be (x, y)
Then AB midpoint D is (x / 2, Y / 2)
So the length of the center line CD on the edge of AB is √ [(4-x / 2) ^ 2 + (0-y / 2) ^ 2] = 3
So (x / 2-4) ^ 2 + y ^ 2 / 4 = 9
(x-8)^2+y^2=36
ABC can't be in a straight line, so y is not equal to 0
If y = 0, then (X-8) ^ 2 = 36, x = 14, x = 2
So the trajectory equation is (X-8) ^ 2 + y ^ 2 = 36, but does not include (14,0), (2,0)

Solving the trajectory equation: given that the BC side length of the triangle ABC is 6 and the perimeter is 16, find the trajectory equation of the vertex ad

AB+AC=16-6=10
So the locus of point a is ellipse and the two intersections with BC are removed
If the midpoint of BC is taken as the coordinate origin, B (-3,0), C (3,0), then there are: x^2/25+y^2/16=1, (y is not equal to 0)

The coordinates of the two vertices a and B of the triangle ABC are (- 6,0), (6,0), respectively. The product of the slope of the straight line where the edges AC and BC are located is equal to − 4 9. Find the trajectory equation of vertex C and draw a sketch

Let the coordinates of vertex C be (x, y), and y can be known from the meaning of the topic
x+6•y
x−6＝−4
9，
The results are as follows: x2
36+y2
16＝1，
When y = 0, point C and point a coincide with point B
Therefore, the trajectory equation of point C is x2
36+y2
16＝1（y≠0），
The sketch is as follows:

Given that the length of edge ab of triangle ABC is 2a, if the median line of BC is fixed length m, then the trajectory equation of vertex C is obtained

Let a (- A, 0), B (a, 0), C (x, y)
Let the midpoint of BC be D, then d ((x + a) / 2, Y / 2)
Since ad = m, i.e. ((a + x) / 2 - (- a)) ^ 2 + (Y / 2-0) ^ 2 = m ^ 2, the equation is obtained

Given that B and C are two fixed points, BC = 6, and the circumference of triangle ABC is equal to 16, find out the path path of vertex a of triangle ABC

Then B (- 3,0), C (3,0). From the meaning of the title (the circumference of the triangle ABC is equal to 16), we can see that ab + AC = 16 - BC = 10, so it is not difficult to find that the locus of point a should be an ellipse. If 2A = 10, a = 5; 2C = 6, C = 3, then B ^ 2 = a ^ 2 - C ^ 2 = 16

In △ ABC, given the vertex a (1,1), B (3,6) and the area of △ ABC is equal to 3, the trajectory equation of vertex C is obtained

Let the coordinates of vertex C be (x, y), and make ch ⊥ AB in H
S=1
2|AB|•|CH|＝3… (2 points)
∵kAB=6−1
3−1＝5
2．
The equation of line AB is Y-1 = 5
2 (x-1), that is, 5x-2y-3 = 0 (4 points)
∴|CH|=|5x−2y−3|
52+(−2)2＝|5x−2y−3|
29… (6 points)
∵|AB|=
(3−1)2+(6−1)2=
29，
∴1
2 x
29×|5x−2y−3|
29＝3… (9 points)
Simplify, get |5x-2y-3|=6, that is, 5x-2y-9=0 or 5x-2y+3=0, which is the trajectory equation of the vertex C (12 points)

In △ ABC, given the vertex a (1,1), B (3,6) and the area of △ ABC is equal to 3, the trajectory equation of vertex C is obtained

Given that B and C are two fixed points, | BC | = 6, and the circumference of △ ABC is equal to 16, the trajectory equation of vertex A is obtained

Taking the straight line where BC is located as the x-axis and the vertical line of BC as the y-axis, a rectangular coordinate system is established, and the vertex a (x, y) can be obtained from the known: | ab | + | AC | = 10 | 6 = | BC |,
According to the definition of ellipse, the locus of point a is an ellipse (excluding the two ends of the long axis), where a = 5, C = 3, B = 4
The standard equation of ellipse is x2
25+y2
16＝1(y≠0)．

In the triangle ABC, we know that the absolute value of BC is equal to 2, the absolute value of AB divided by the absolute value of AC is equal to m, and the trajectory equation of point a is obtained It must be answered on Sunday

Let BC be placed horizontally, B on the left, O ∈ BC, so that Bo ∶ OC = m ∶ 1
Take o as the origin, OC as the x-axis, with the y-axis. There are B (- 2m / (M + 1), 0), C (2 / (M + 1), 0)
Let a (x.y). ∵ (│ ab │) / (│ AC │) = M
∴[（x+2m／（m+1））²+y²]／[（x-2／（m+1））²+y²]＝m².
The results are as follows: 1
When m ≠ 1: [x + 2m / (1-m 2)] 2 + y 2 = [2 m / (1-m 2)] 2
This is a circle with a center (- 2m / (1-m?), 0) and a radius of | 2m / (1-m |)
When m = 1: x = 0
The vertical bisector of BC

Given that B and C are two fixed points, the absolute value BC is equal to 8, and the circumference of triangle ABC is equal to 18, the trajectory equation of vertex A is obtained The result is (x ^ 2) / 25 + (y ^ 2) / 9 = 1 But the answer is (Y ≠ 0) Why can't y be equal to 0

When y = 0, point a is (- 5,0) or (5,0) on the coordinate axis, which can not form a triangle, so y cannot be equal to 0. The key to doing mathematical problems is to analyze the known conditions of a topic. The same is true for college entrance examination. See the meaning of the topic and then do it

The coordinates of the vertex B and C of the triangle ABC are (0,0) and (4,0) AB, and the length of the midline on the edge of (4,0) AB is 3