Two congruent triangle plates ade and ABC with 30 ° and 60 ° angles are placed as shown in the figure. E, a and C are in a straight line to connect BD. take the midpoint m of BD to connect me and MC. Try to judge the shape of △ EMC and explain the reason

Two congruent triangle plates ade and ABC with 30 ° and 60 ° angles are placed as shown in the figure. E, a and C are in a straight line to connect BD. take the midpoint m of BD to connect me and MC. Try to judge the shape of △ EMC and explain the reason

The reasons are as follows: connecting ma. ∵ ead = 30 °, BAC = 60 °, ∵ DAB = 90 °, ∵ EDA ≌ cab, ∵ Da = AB, ed = AC, ∵ DAB are isosceles right triangles. And ∵ m is the midpoint of BD, ∵ MDA = ∠ MBA = 45 °, am ⊥ BD (three in one), am = 12

2x+5y=1/2,3x-5y=1/3
There is another way: X / 2 + Z / 3 = 7, X / 3-z / 4 = - 1 / / / / / / / use addition and subtraction

2X + 5Y = 1 / 2 is in duplicate
3x-5y = 1 / 3 is a two formula
One + two, get:
5X = 5/6
X=1/6
Substituting x = 1 / 6 into the original,
2 * 1/6 + 5Y = 1/2
Y =1/30
If x / 2 + Z / 3 = 7 is in duplicate and both sides of the equal sign multiply by 3, the three formulas 3x / 2 + Z = 21 are obtained
X / 3-z / 4 = - 1 is a two formula, and the two sides of the equal sign multiply by 4 to get the four formula 4x / 3-z = - 4
Three plus four, get:
17X / 6 = 17
X = 6
Substituting x = 6 into the original, we get,
6/2 + Z/3 = 7
Z = 12

2x-6y = 1 2x + y = - 4
3 / 2x-6 / 5Y = 1, 2x + y = - 4 to solve the equation

2X of 3-5Y of 6 = 1 ①
2x+y=-4②
The equation (1) is reduced to 4x-5y = 6 (3)
② Y = - 2
Substituting y = - 2 into 2, we get x = - 1
The solution of the equations is x = - 1
y=-2

Given three points a (1,3), B (- 2,0), C (2,4), judge whether these three points are on the same straight line, and explain the reason

These three points are on the same straight line. The reason is as follows: let the analytic formula of the straight line passing through point a and point B be y = KX + B, substitute a (1,3), B (- 2,0) into K + B = 3 − 2K + B = 0, and solve k = 1b = 2, so the analytic formula of the straight line AB is y = x + 2, when x = 2, y = 2 + 2 = 4, so point C (2,4) is on the straight line y = x + 2, that is, the three points are on the same straight line

Given three points a (1,3), B (- 2,4), C (2,4), try to judge whether these three points are on the same straight line, and explain the reason

The solution of the equation: Y1 = KX + B 3 = K + B 4 = - 2K + B is: k = - 1 / 3, B = 10 / 3, then Y1 = - 1 / 3x + 10 / 3, the solution of the equation: y2 = KX + B 4 = - 2K + B 4 = 2K + B is: k = 0, B = 4, y2 = 4, so the three points are not on the same line

A (- 1, - 1), B (1,3) C (2,5) are known. Judge the position of a.b.c

Let y = ax + B
The results are as follows:
-a+b=-1
a+b=3
Solution
a=2
b=1
Three points on the line y = 2x + 1

It is known that a (- 1, - 1) B (1,3) C (2,5) can be used to judge the position relationship of ABC

K1=(3+1)/(1+1)=2,
K2=(5-3)/(2-1)=2
K1=K2
A (- 1, - 1) B (1,3) C (2,5) on a line y = 2x + 1

If a (13, 1a), B (14, 1b), C (15, 1c) satisfy AB + C = 13, Ba + C = 12, then the positions of a, B, C are suitable for ()
A. Form an acute triangle B. form a right triangle C. form an obtuse triangle D. on the same line

∵ AB + C = 13, ∵ 3A = B + C & nbsp; & nbsp; & nbsp; ① and ∵ Ba + C = 12, ∵ 2B = a + C ②. From ① and ②, we get b = 43a, C = 53a, ∵ a (13, 1a), B (14, 34a), C (15, 35a). Let the analytic expression of the straight line AB be y = KX + B. substituting the coordinates of points a and B, we get k = 3A, B = 0,